8.1: Plain View Doctrine
“It is well established that under certain circumstances the police may seize evidence in plain view without a warrant.” 1
here is general agreement that the plain view rule is fairly simple to understand and apply. Even the words “plain view” seem to to saying,
“If it’s visible, it’s seizable!” Of course, it is not that simple, but it’s not very complicated either. Specifically, evidence is deemed in plain view—and can therefore be seized without a warrant—if the following circumstances existed:
- Lawful vantage point: The officers’ initial viewing of the evidence must have been “lawful.”
- Probable cause: Before seizing the evidence, officers must have had probable cause to believe it was, in fact, evidence of a crime
- Lawful access: Officers must have had a legal right to enter the place in which the evidence was located.
If these circumstances exist, the officers’ act of observing the evidence does not constitute a “search” because no one can reasonably expect privacy in something that is so readily exposed; and their act of seizing the evidence is lawful because the plain view rule constitutes an exception to the warrant requirement. 2 As the United States Supreme Court explained, “The seizure of property in plain view involves no invasion of privacy and is presumptively reasonable.” 3
Lawful Vantage Point
The requirement that the officers’ initial observation of the evidence must have been “lawful” is satisfied if the officers did not violate the suspect’s Fourth Amendment rights by getting into the position from which they saw it. 4 “The plain view doctrine,” said the Supreme Court, “is grounded on the proposition that once police are lawfully in a position to observe an item first-hand, its owner’s privacy interest in that item is lost.” 5
Before we discuss the types of places from which an observation is apt to be legal, it should be noted an observation does not become an unlawful search merely because officers had to make some effort to see the evidence, so long as the effort was reasonably foreseeable. Thus, it is unimportant that officers could not initially see the evidence without using a common visual aid (such as a flashlight or binoculars), 6 or without bending down or elevating themselves somewhat. Thus, the D.C. Circuit explained, “That a policeman may have to crane his neck, or bend over, or squat, does not render the [plain view] doctrine inapplicable, so long as what he saw would have been visible to any curious passerby.” 7 Similarly, the Court of Appeal ruled that merely looking over the five-foot fence from a neighbor’s yard “disclosed no more than what was in plain view.” 8
In contrast, the courts have ruled that officers “searched” a high-rise apartment when they could only see the evidence inside by using high-power binoculars from a hilltop about 250 yards away, 9 or when officers “had to squeeze into a narrow area between the neighbor’s garage and defendant’s fence” and that area was almost blocked by foliage. 10
OBSERVATION FROM PUBLIC PLACE
The most obvious example of a lawful vantage point is a place that is accessible to the general public. 11 Thus, the Supreme Court pointed out that “the police may see what may be seen from a public vantage point where they have a right to be,” 12 and that officers “cannot reasonably be expected to avert their eyes from evidence of criminal activity that could have been observed by any member of the public.” 13
OBSERVATION DURING DETENTION OR ARREST
An observation that occurred in the course of a detention is lawful if officers had sufficient grounds for the detention or arrest and it was reasonable in its scope and intensity. 14 For example, in People v. Sandoval 15 the Court of Appeal ruled that an officer, having made a lawful car stop, lawfully observed drugs and paraphernalia in the passenger compartment because “the officer clearly had a right to be in the position to have that view.”
OBSERVATION DURING PAT SEARCH
In a variation of the plain view rule (i.e., the “plain feel” rule), officers who feel evidence while conducting a pat search are deemed to be in a lawful vantage point if they had grounds for the search. 16 In such cases, said the Third Circuit, the “proper question” is whether the officer detected the evidence “in a manner consistent with a routine frisk.” 17 Or, in the words of the Supreme Court, a lawful pat search must “be confined in scope to an intrusion reasonably designed to discover guns, knives, clubs, or other hidden instruments for the assault of the police officer.” 18
OBSERVATION WHILE EXECUTING A SEARCH WARRANT
Officers who are executing search warrants often find evidence that was not listed in the warrant. When this happens, the discovery will be deemed lawful under the plain view rule if they found the evidence while looking in places or things in which any of the listed evidence might have been found. For example, in Skelton v. Superior Court 19 officers in La Palma were searching for a wedding ring and carving set which were taken in a burglary. While searching for these items, they also found some watches and rings that matched the descriptions of items taken in related burglaries. On appeal, the California Supreme Court ruled the unlisted evidence was lawfully discovered because “the warrant mandated a search for and seizure of several small and easily secreted items” and therefore “the officers had the authority to conduct an intensive search of the entire house.”
Similarly, in U.S. v. Smith , 20 officers in Tampa obtained a warrant to search the home of Smith’s mother for drugs and indicia. In the course of the search, they opened Smith’s lockbox and found child pornography. In ruling that the pornography was discovered lawfully, the court said, “It was through the lawful execution of the warrant that the officers came across the photographs at issue here.”
In contrast, in People v. Albritton 21 narcotics officers in Bakersfield obtained a warrant to search the defendant’s home for drugs and indicia. A detective assigned to the auto theft detail learned about the warrant and decided to “go along for the ride” because the defendant was also a suspected car thief. When the officers arrived, the detective “immediately separated himself from the others and went to the garage” where he checked the VIN numbers on several cars and learned that four were stolen. On appeal, prosecutors argued that the detective’s initial viewing of the VIN numbers was lawful, and therefore the plain view rule applied. But the court disagreed, ruling the detective’s observation of the VIN numbers was unlawful because none of the evidence listed in the search warrant could reasonably have been found in the areas in which the VIN numbers were located.
OBSERVATION DURING WARRANTLESS ENTRY
In a similar vein, officers may seize evidence inside a residence if (1) they were lawfully on the premises (e.g., exigent circumstances, consentual entry, execution of an arrest warrant), and (2) they discovered the evidence while they were carrying out their lawful duties. For example, if the officers’ entry into a living room was consensual (e.g., a knock and talk), and if they saw drugs in the room, their observation would be deemed lawful because they had been invited into that room. But if they saw the evidence by opening a container in the living room or while wandering into another room, the observations would be unlawful.
A good example of such an unlawful observation is found in Arizona v. Hicks 22 in which officers had entered Hicks’ apartment without a warrant because someone in his apartment had fired a shot through the floor, injuring an occupant in the apartment below. While looking around, one of the officers noticed an expensive audio system which he thought might have been stolen because the apartment was otherwise “squalid.” The officer then confirmed his suspicion by picking up a component, writing down the serial number, and running it through a police database. Although the U.S. Supreme Court ruled that the entry into the apartment was lawful, it ruled that the serial number was not in plain view because the officer could not have seen it without doing something (picking up the component) that went beyond the objective of the entry, which was to apprehend the shooter and look for any other injured people.
OBSERVATION DURING ENTRY INTO YARDS
As with warrantless entries into residences, warrantless entries into a suspect’s front, back, or side yards may fall within an exception to the warrant requirement (e.g., exigent circumstances, consent), in which case their observations would be lawful. In the absence of a warrant, officers may still walk to the front door via normal access routes, then knock or otherwise announce their presence. But if no one answers the door within a reasonable time, any observations they make may be illegal if they loitered on the property or explored the grounds. As the Supreme Court explained, officers are impliedly authorized “to approach the home by the front path, knock promptly, wait briefly to be received, and then (absent invitation to linger longer) leave.” 23
For example, in People v.  A sheriff’s deputy who was investigating the murder drove to Madera and, while standing on Edelbacher’s driveway, saw shoe prints that looked just like the shoe prints that had been found at the murder scene. Consequently, officers took photos of the shoe prints and prosecutors used them against Edelbacher at his trial. On appeal, he argued that the discovery was unlawful because the deputy had been standing on his private property. It didn’t matter, said the California Supreme Court, because the prints “were apparently visible on the normal route used by visitors approaching the front doors of the residences and there is no indication of solid fencing or visible efforts to establish a zone of privacy.”
OBSERVATION FROM ADJACENT PROPERTY
An observation of evidence in a suspect’s yard or other private property is not unlawful if it was made from a neighbor’s property, even if the officers were technically trespassing. 25 This is because it was the neighbor who was intruded upon—not the suspect. As the Court of Appeal observed, “[A] search does not violate the Fourth Amendment simply because police officers trespassed onto a neighbor’s property when making their observations.” 26
OBSERVATION DURING COMPUTER SEARCH
Officers who are executing a warrant to search a computer will often discover unlisted data or evidence of some other crime. When this happens the discovery will be deemed lawful under the plain view rule if the file in which the evidence was found could have contained any of the data or graphics listed in the warrant. In most cases, that means every file must be read because, as the Ninth Circuit pointed out in U.S. v. Comprehensive Drug Testing, Inc. , unless officers read every file they would have “no way of knowing which or how many illicit files there might be or where they might be stored.” 27
Probable Cause
The second requirement for a plain view seizure is that the officers—at or before the moment they seized the evidence—must have had probable cause to believe the item was, in fact, evidence of a crime. 28 And like the other forms of proof, probable cause to seize an item in plain view may be based on direct or circumstantial proof. Examples of direct proof would include an officer’s observation of a weapon that is illegal to possess, 29 a weapon used in a crime, 30 readily-identifiable drugs or drug paraphernalia, 31 readily-identifiable child pornography, 32 or property that had been reported stolen. 33
As we will now discuss, circumstantial proof typically consists of an officer’s observation of something that, based on his training and experience, appears to be seizable evidence.
INSTRUMENTALITIES OF A CRIME
Probable cause is often based on an officer’s knowledge of a link between the item and a certain crime or a type of crime.  For example, in People v. Gorak 47 the court ruled that officers had probable cause to seize an air compressor in plain view in the back seat of the defendant’s car mainly because “the electrical lines and air lines appeared to have been broken off” and water was leaking out of a broken line. Similarly, in People v. Stokes 48 two Hayward police officers in an unmarked car were driving through a mobile home park that was occupied mainly by senior citizens. As they turned a corner, they saw Stokes standing in the middle of the street, holding a video recorder. The officers recognized Stokes as a local burglar, they noticed that he kept looking around and appeared to be nervous, that he was carrying a screwdriver, and that several homes in the park had recently been burglarized. Although the officers had no direct evidence that the recorder had been stolen, the court ruled that the circumstantial evidence was quite sufficient.
Other circumstantial evidence that may suffice include the presence of store merchandise tags or anti-shoplifting devices that are usually removed when retail goods are sold; or the presence of an inordinate amount of property, especially the type of property that is frequently stolen, such as TVs, cell phones, tablets, firearms, and jewelry. 49
POSSESSION OF DRUGS, PARAPHERNALIA
Officers frequently develop probable cause to seize a container in the possession of a drug user or trafficker based entirely on circumstantial evidence that it contained drugs, paraphernalia, or evidence of sales. 50 As the court observed in People v. Holt , “Courts have recognized certain containers as distinctive drug carrying devices which may be seized upon observation: heroin balloons, paper bindles and marijuana smelling brick-shaped packages.” 51
Probable cause may also be based on how the object felt; i.e., “plain feel.” 52 For example, in People v. Lee 53 an Oakland police officer was pat searching a suspected drug dealer when he felt “a clump of small resilient objects” which he believed (correctly) were heroin-filled balloons. In ruling that the officer’s seizure of the balloons was lawful under the “plain feel” rule, the court noted that he “recognized the feel of such balloons from at least 100 other occasions on which he had pat-searched people and felt what were later determined to be heroin-filled balloons. As he described it, the feel is unmistakable.”
Lawful Access
Finally, even if officers could see the evidence and had probable cause to believe it was seizable, they may not enter the suspect’s home or other place in which he had a reasonable expectation of privacy unless they had a legal right to enter; e.g., a vehicle in which the evidence was located. 54 Thus, in discussing the plain view rule, the Supreme Court explained that “not only must the officer be lawfully located in a place from which the object can be plainly seen, but he or she must also have a lawful right of access to the object itself.” 55 Or, as Justice Grodin observed in People v. Superior Court (Spielman) , “Seeing something in plain view does not, of course, dispose, ipso facto, of the problem of crossing constitutionally protected thresholds. Those who thoughtlessly overapply the plain view doctrine to every situation where there is a visual open view have not yet learned the simple lesson long since mastered by old hands at the burlesque houses, ‘You can’t touch everything you can see.’” 56 Note that officers will always have lawful access to evidence located in a public place or a vehicle located in a public place. 57 In addition, they may enter a residence and seize evidence observed from the outside if they were aware that a resident was subject to a parole or probation search or if they reasonably believed the evidence would be destroyed if they delayed seizing it.
For example, in People v. Ortiz 58 an officer happened to be walking by the open door of a hotel room when he saw a woman inside, and she was “counting out tinfoil bindles and placing them on a table.” Having probable cause to believe the bindles contained heroin, the officer went inside, seized the bindles, and arrested the woman and the other occupants. In ruling that the officer had lawful access to the evidence, the court pointed out that, because he was initially only three to six feet away from the woman, he reasonably believed that she had seen him and it is “common knowledge that those who possess drugs often attempt to destroy the evidence when they are observed by law enforcement officers.” Consequently, the court ruled that the officer had a legal right to enter because “it was reasonable for [him] to believe the contraband he saw in front of defendant and the woman was in imminent danger of being destroyed.
References
- Coolidge v. New Hampshire (1971) 403 U.S. 443, 465.
- See People v. Kilpatrick (1980) 105 Cal.App.3d 401, 408; People v. Albritton (1982) 138 Cal.App.3d 79, 85, fn.1.
- Payton v. New York (1980) 445 U.S. 573, 587.
- See Minnesota v. Dickerson (1993) 508 U.S. 366, 375 [“The rationale of the plain-view doctrine is that if contraband is left in open view and is observed by a police officer from a lawful vantage point, there has been no invasion of a legitimate expectation of privacy and thus no ‘search’”]; Washington v. Chrisman (1982) 455 U.S. 1, 5-6 [“The ‘plain view’ exception to the Fourth Amendment warrant requirement permits a law enforcement officer to seize what clearly is incriminating evidence or contraband when it is discovered in a place where the officer has a right to be.”]; People v. Bradford (1997) 15 Cal.4th 1229, 1295.
- Illinois v. Andreas (1983) 463 U.S. 765, 771.
- See On Lee v. United States (1952) 343 U.S. 747, 754; Texas v. Brown (1983) 460 U.S. 730, 740; People v. Superior Court (Mata) (1970) 3 Cal.App.3d 636, 639; People v. St. Amour (1980) 104 Cal.App.3d 886, 893 [“So long as the object which is viewed is perceptible to the naked eye … the government may use technological aid of whatever type without infringing on the person’s Fourth Amendment rights.”].
- James v. U.S. (D.C. Cir. 1969) 418 F.2d 1150, 1151.
- People v. Superior Court (Stroud) (1974) 37 Cal.App.3d 836, 839.
- People v. Arno (1979) 90 Cal.App.3d 505. Also see People v. Henderson (1990) 220 Cal.App.3d 1632, 1649.
- People v. Fly (1973) 34 Cal.App.3d 665, 667. Also see Pate v. Municipal Court (1970) 11 Cal.App.3d 721, 724 [officer climbed over a fence onto a trellis, then walked along the trellis for a considerable distance]; Jacobs v. Superior Court (1973) 36 Cal.App.3d 489 [the officer had to step onto a small planter area between the building and the parking lot]; Lorenzana v. Superior Court (1973) 9 Cal.3d 626, 636 [officer had to traverse some bushes that constituted a “significant hindrance”].
- See Katz v. United States (1967) 389 U.S. 347, 351 [“What a person knowingly exposes to the public, even in his own home or office, is not a subject of Fourth Amendment protection.”]; Florida v. Riley (1989) 488 U.S. 445, 449-50 [“Thus the police, like the public, would have been free to inspect the backyard garden from the street if their view had been unobstructed.”]; People v. Deutsch (1996) 44 Cal.App.4th 1224, 1229 [“Information or activities which are exposed to public view cannot be characterized as something in which a person has a subjective expectation of privacy.”].
- Florida v. Riley (1989) 488 U.S. 445, 449.
- California v. Greenwood (1988) 486 U.S. 35, 41.
- See United States v. Hensley (1985) 469 U.S. 221, 235; Texas v. Brown (1983) 460 U.S. 730, 737, 739; People v. DeCosse (1986) 183 Cal.App.3d 404, 410 [“Standing where he had a right to be, the officer was lawfully entitled to observe, in plain sight, the opened alcoholic beverage container.”].
- (1985) 164 Cal.App.3d 958.
- See People v. Avila (1997) 58 Cal.App.4th 1069, 1075 [“However, if contraband is found while performing a permissible Terry search, the officer cannot be expected to ignore that contraband.”]; People v. Armenta (1968) 268 Cal.App.2d 248, 253 [“The officer was not required to blind himself to the heroin simply because it was disconnected from the initial purpose of the search.”]; People v. Garcia (1969) 274 Cal.App.2d 100, 106-7 [“[T]he manner of conducting an otherwise justified precautionary search is of vital importance.”].
- U.S. v. Yamba (3rd Cir. 2007) 506 F.3d 251, 259.
- Terry v. Ohio (1968) 392 U.S. 1 29.
- (1969) 1 Cal.3d 144. Also see Horton v. California (1990) 496 U.S. 128, 142.
- (11th Cir. 2006) 459 F.3d 1276.
- (1982) 138 Cal.App.3d 79.
- (1987) 480 U.S. 321.
- Florida v. Jardines (2013) U.S. [133 S.Ct. 1409, 1415].
- (1989) 47 Cal.3d 983.
- See Dillon v. Superior Court (1972) 7 Cal.3d 305, 311 [officer’s observation of a marijuana garden in a fenced-in backyard was lawful where the officer viewed the garden from the second floor of the house next door whose owner had consented to the entry]; People v. Shaw (2002) 97 Cal.App.4th 833 [with permission of a neighbor, officers standing behind a fence looked into the common area of defendant’s apartment]; People v. Smith (1986) 180 Cal.App.3d 72, 83-84 [“The fence surrounding Smith’s (marijuana) garden was only five feet high and allowed people outside to see the activities occurring inside the garden.”].
- People v. Claeys (2002) 97 Cal.App.4th 55, 59.
- (9th Cir. 2010) 621 F.3d 1162, 1171. Also see U.S. v. Schesso (9th Cir. 2013) 730 F.3d 1040, 1046.
- (7th Cir. 2010) 592 F.3d 779, 785. Also see U.S. v. Stabile (3rd Cir. 2011) 633 F.3d 219, 239 [“Detective Vanadia’s decision to highlight and view the contents of the Kazvid folder was objectively reasonable because criminals can easily alter file names and file extensions to conceal contraband.”].
- See Arizona v. Hicks (1987) 480 U.S. 321, 326; People v. Stokes (1990) 224 Cal.App.3d 715, 719. NOTE: In Coolidge v. New Hampshire (1971) 403 U.S. 443, 466 a plurality of the Supreme Court said that officers may not seize evidence in plain view unless it was “immediately apparent” that the item was evidence of a crime. Subsequently, the Court observed that the term “immediately apparent” was “very likely an unhappy choice of words, since it can be taken to imply that an unduly high degree of certainty as to the incriminatory character of evidence is necessary for an application of the ‘plain view’ doctrine.” Texas v. Brown (1983) 460 U.S. 730, 741. The Court then ruled that only probable cause is required. At p. 742. Also see Minnesota v. Dickerson (1993) 508 U.S. 366, 375; People v. Clark (1989) 212 Cal.App.3d 1233, 1238.
- Texas v. Brown (1983) 460 U.S. 730, 742. Also see People v. Stokes (1990) 224 Cal.App.3d 715, 719.
- See People v. McNeal (1979) 90 CA3 830, 841 [nunchucks].
- Horton v. California (1990) 496 U.S. 128, 131 [stun gun used in robbery]; People v. Bradford (1997) 15 Cal.4th 1229, 1296.
- See People v. Nickles (1970) 9 Cal.App.3d 986, 994; People v. LeBlank (1997) 60 Cal.App.4th 157, 165.
- See U.S. v. Benoit (10th Cir.2013) 713 F.3d 1, 11.
- See Colorado v. Bannister (1980) 449 U.S. 1, 2; Christians v. Chester (1990) 218 Cal.App.3d 273, 275.
- U.S. v. Muhammad (8th Cir. 2010) 604 F.3d 1022, 1027-28.
- See People v. Rico (1979) 97 Cal.App.3d 124, 133; People v. Superior Court (Orozco) (1981) 121 CA3 395, 404,
- Warden v. Hayden (1967) 387 U.S. 294.
- People v. Jardine (1981) 116 Cal.App.3d 907, 913.
- Horton v. California (1990) 496 U.S. 128, 130-1, 142.
- People v. Easley (1983) 34 Cal.3d 858, 872.
- People v. Hill (1974) 12 Cal.3d 731, 763.
- People v. Miley (1984) 158 Cal.App.3d 25, 35-36.
- People v. Vasquez (1983) 138 Cal.App.3d 995, 999-1000.
- People v. Koelzer (1963) 222 Cal.App.3d 20, 25; People v. Mack (1977) 66 Cal.App.3d 839, 859.
- People v. Taylor (1975) 46 Cal.App.3d 513, 518.
- (1987) 196 Cal.App.3d 1032.
- (1990) 224 Cal.App.3d 715.
- See In re Donald L. (1978) 81 Cal.App.3d 770, 775 [the officer “could have reasonably believed that the assorted objects of jewelry, including women’s jewelry, were probably stolen”]; In re Curtis T. (1989) 214 Cal.App.3d 1391, 1398; People v. Sedillo (1982) 135 Cal.App.3d 616, 623; People v. Williams (1988) 198 Cal.App.3d 873, 890; People v. McGraw (1981) 119 Cal.App.3d 582, 603; People v. Atkins (1982) 128 Cal.App.3d 564, 570; People v. Garcia (1981) 121 Cal.App.3d 239, 246; People v. Superior Court (Thomas) (1970) 9 Cal.App.3d 203, 210; People v. Jennings (1965) 231 Cal.App.2d 744.
- See Texas v. Brown (1983) 460 U.S. 730, 743 [“[T]he distinctive character of the balloon itself spoke volumes as to its contents— particularly to the trained eye of the officer.”]; United States v. Jacobsen (1984) 466 U.S. 109, 121 [“it was just like a balloon the distinctive character of which spoke volumes as to its contents”]; People v. Nonnette (1990) 221 Cal.App3d 659, 666 [bundle of tiny baggies of the type used for drugs]; People v. Chapman (1990) 224 Cal.App.3d 253, 257 [“Probable cause to believe a container holds contraband may be adequately afforded by its shape, design, and the manner in which it is carried.
- 212 Cal.App.3d 1200, 1205.
- See People v. Dibb (1995) 37 Cal.App.4th 832, 836-37 [“The critical question is not whether [the officer] could identify the object as contraband based on only the ‘plain feel’ of the object, but whether the totality of circumstances made it immediately apparent to [the officer] when he first felt the lump that the object was contraband.”]; People v. Chavers (1983) 33 Cal.3d 462, 471 [“[T]he knowledge [gained by the officer through sense of touch] was as meaningful and accurate as if the container had been transparent and he had seen the gun within the container.”].
- (1987) 194 Cal.App.3d 975.
- See Texas v. Brown (1983) 460 U.S. 730, 738 [“[P]lain view provides grounds for seizure of an item when an officer’s access to an object has some prior justification under the Fourth Amendment.”]; United States v. Ross (1982) 456 U.S. 798, 809; People v. Ortiz (1995) 32 Cal.App.4th 286, 291 [“Before Officer Forsythe could enter the hotel room to arrest defendant and seize the tinfoil bindles containing heroin, he needed to have a lawful right of access to defendant and the heroin.”]; U.S. v. Davis (4th Cir. 2012) 690 F.3d 226, 234 [“the lawful access requirement is intended to clarify that police may not enter a premises to make a warrantless seizure, even if they could otherwise see (from a lawful vantage point) that there was contraband in plain sight”].
- Horton v. California (1990) 496 U.S. 128, 137.
- (1980) 102 Cal.App.3d 342, 348, fn.1 (conc. opn. Grodin, J.).
- See United States v. Ross (1982) 456 U.S. 798, 809; People v. Superior Court (Nasmeh) (2007) 151 Cal.App.4th 85, 100.
- (1995) 32 Cal.App.4th 286.


1.3: Introduction to Combustion Analysis
Applications of Combustion Analysis
Combustion, or burning as it is more commonly known, is simply the mixing and exothermic reaction of a fuel and an oxidizer. It has been used since prehistoric times in a variety of ways, such as a source of direct heat, as in furnaces, boilers, stoves, and metal forming, or in piston engines, gas turbines, jet engines, rocket engines, guns, and explosives. Automobile engines use internal combustion in order to convert chemical into mechanical energy. Combustion is currently utilized in the production of large quantities of \(\ce{H2}\). Coal or coke is combusted at 1000 ◦ C in the presence of water in a two-step reaction. The first step shown in involved the partial oxidation of carbon to carbon monoxide.
\[\ce{C(g) + H2O(g) -> CO(g) + H2(g)} \nonumber \]
The second step involves a mixture of produced carbon monoxide with water to produce hydrogen and is commonly known as the water gas shift reaction.
\[\ce{CO(g) + H2O(g) → CO2(g) + H2(g)} \nonumber \]
Although combustion provides a multitude of uses, it was not employed as a scientific analytical tool until the late 18 th century.
History of Combustion
In the 1780's, Antoine Lavoisier (figure \(\PageIndex{1}\) ) was the first to analyze organic compounds with combustion using an extremely large and expensive apparatus (figure \(\PageIndex{2}\) ) that required over 50 g of the organic sample and a team of operators.
The method was simplified and optimized throughout the 19 th and 20 th centuries, first by Joseph Gay- Lussac (Figure \(\PageIndex{3}\)), who began to use copper oxide in 1815, which is still used as the standard catalyst.
William Prout (Figure \(\PageIndex{4}\)) invented a new method of combustion analysis in 1827 by heating a mixture of the sample and \(\ce{CuO }\) using a multiple-flame alcohol lamp (Figure \(\PageIndex{5}\)) and measuring the change in gaseous volume.
In 1831, Justus von Liebig (Figure \(\PageIndex{6}\))) simplified the method of combustion analysis into a "combustion train" system (Figure \(\PageIndex{7}\)) and Figure \(\PageIndex{8}\))) that linearly heated the sample using coal, absorbed water using calcium chloride, and absorbed carbon dioxide using potash (KOH). This new method only required 0.5 g of sample and a single operator, and Liebig moved the sample through the apparatus by sucking on an opening at the far right end of the apparatus.
Jean-Baptiste André Dumas (Figure \(\PageIndex{9}\))) used a similar combustion train to Liebig. However, he added a U-shaped aspirator that prevented atmospheric moisture from entering the apparatus (Figure \(\PageIndex{10}\))).
In 1923, Fritz Pregl (Figure \(\PageIndex{11}\))) received the Nobel Prize for inventing a micro-analysis method of combustion. This method required only 5 mg or less, which is 0.01% of the amount required in Lavoisier's apparatus.
Today, combustion analysis of an organic or organometallic compound only requires about 2 mg of sample. Although this method of analysis destroys the sample and is not as sensitive as other techniques, it is still considered a necessity for characterizing an organic compound.
Categories of combustion
Basic flame types
There are several categories of combustion, which can be identified by their flame types (Table \(\PageIndex{1}\)). At some point in the combustion process, the fuel and oxidant must be mixed together. If these are mixed before being burned, the flame type is referred to as a premixed flame, and if they are mixed simultaneously with combustion, it is referred to as a nonpremixed flame. In addition, the ow of the flame can be categorized as either laminar (streamlined) or turbulent (Figure \(\PageIndex{12}\)).
| Fuel/oxidizer mixing | Fluid motion | Examples |
|---|---|---|
| Premixed | Turbulent | Spark-ignited gasoline engine, low NO x stationary gas turbine |
| Premixed | Laminar | Flat flame, Bunsen flame (followed by a nonpremixed candle for Φ>1) |
| Nonpremixed | Turbulent | Pulverized coal combustion, aircraft turbine, diesel engine, H 2 /O 2 rocket motor |
| Nonpremixed | Laminar | Wood fire, radiant burners for heating, candle |
The amount of oxygen in the combustion system can alter the ow of the flame and the appearance. As illustrated in Figure \(\PageIndex{13}\), a flame with no oxygen tends to have a very turbulent flow, while a flame with an excess of oxygen tends to have a laminar flow.
Stoichiometric combustion and calculations
A combustion system is referred to as stoichiometric when all of the fuel and oxidizer are consumed and only carbon dioxide and water are formed. On the other hand, a fuel-rich system has an excess of fuel, and a fuel-lean system has an excess of oxygen (Table \(\PageIndex{2}\)).
| Combustion type | Reaction example |
|---|---|
| Stoichiometric | \(\ce{2H2 + O2 -> 2H2O}\) |
| Fuel-rich (\(\ce{H2}\) left over) | \(\ce{3H2 +O2 ->2H2O+H2}\) |
| Fuel-lean (\(\ce{O2}\) left over) | \(\ce{CH4 +3O2 ->2H2O+CO2 +O2}\) |
If the reaction of a stoichiometric mixture is written to describe the reaction of exactly 1 mol of fuel (\(\ce{H2}\) in this case), then the mole fraction of the fuel content can be easily calculated as follows, where \( ν \) denotes the mole number of \(\ce{O2}\) in the combustion reaction equation for a complete reaction to \(\ce{H2O}\) and \(\ce{CO2}\),
\[ x_{\text{fuel, stoich}} = \dfrac{1}{1+v} \nonumber \]
For example, in the reaction
\[\ce{H2 + 1/2 O2 → H2O2 + H2} \nonumber \]
we have \( v = \frac{1}{2} \), so the stoichiometry is calculated as
\[ x_{\ce{H2}, \text{stoich}}= \dfrac{1}{1+0.5} = 2/3 \nonumber \]
However, as calculated this reaction would be for the reaction in an environment of pure oxygen. On the other hand, air has only 21% oxygen (78% nitrogen, 1% noble gases). Therefore, if air is used as the oxidizer, this must be taken into account in the calculations, i.e.
\[ x_{\ce{N2}} = 3.762 (x_{\ce{O2}}) \nonumber \]
The mole fractions for a stoichiometric mixture in air are therefore calculated in following way:
\[ x_{\text{fuel, stoich}} = \dfrac{1}{1+v(4.762)} \label{eq:xfuel} \]
\[ x_{\ce{O2},\text{stoich}} = v(x_{\text{fuel, stoich}}) \nonumber \]
\[ x_{\ce{N2},\text{stoich}} = 3.762(x_{\ce{O2}, \text{stoich}}) \nonumber \]
Calculate the fuel mole fraction (\(x_{\text{fuel}} \)) for the stoichiometric reaction:
\[ \ce{CH4 + 2O2} + (2 \times 3.762)\ce{N2 → CO2 + 2H2O} + (2 \times 3.762)\ce{N2} \nonumber \]
Solution
In this reaction \( ν \) = 2, as 2 moles of oxygen are needed to fully oxidize methane into \(\ce{H2O}\) and
\(\ce{CO2}\).
\[ x_{\text{fuel, stoich}} = \dfrac{1}{1+2 \times 4.762} = 0.09502 = 9.502~\text{mol} \% \nonumber \]
Calculate the fuel mole fraction for the stoichiometric reaction:
\[ \ce{C3H8 + 5O2} + (5 \times 3.762)\ce{N2 → 3CO2 + 4H2O} + (5 \times 3.762)\ce{N2} \nonumber \]
- Answer
-
The fuel mole fraction is 4.03%
Premixed combustion reactions can also be characterized by the air equivalence ratio, \( \lambda \):
\[ \lambda = \dfrac{x_{\text{air}}/x_{\text{fuel}}}{x_{\text{air, stoich}}/x_{\text{fuel,stoich}}} \nonumber \]
The fuel equivalence ratio, \( Φ\), is the reciprocal of this value
\[ Φ = 1/\lambda \nonumber \]
Rewriting \ref{eq:xfuel} in terms of the fuel equivalence ratio gives:
\[ x_{\text{fuel}} = \frac { 1 } { 1 + v( 4.672 / \Phi ) } \nonumber \]
\[ x_{\text{air}} = 1 - x_{\text{fuel}} \nonumber \]
\[ x_{\ce{O2}} = x_{\text{air}}/4.762 \nonumber \]
\[ x_{\ce{N2}} = 3.762(x_{\ce{O2}}) \nonumber \]
The premixed combustion processes can also be identified by their air and fuel equivalence ratios (Table \(\PageIndex{3}\) ).
| Type of combustion | Φ | λ |
|---|---|---|
| Rich | >1 | <1 |
| Stoichiometric | =1 | =1 |
| Lean | <1 | >1 |
With a premixed type of combustion, there is much greater control over the reaction. If performed at lean conditions, then high temperatures, the pollutant nitric oxide, and the production of soot can be minimized or even avoided, allowing the system to combust efficiently. However, a premixed system requires large volumes of premixed reactants, which pose a fire hazard. As a result, nonpremixed combusted, while not being efficient, is more commonly used.
Instrumentation
Though the instrumentation of combustion analysis has greatly improved, the basic components of the apparatus (Figure 1.14) have not changed much since the late 18 th century.
The sample of an organic compound, such as a hydrocarbon, is contained within a furnace or exposed to a ame and burned in the presence of oxygen, creating water vapor and carbon dioxide gas (Figure \(\PageIndex{15}\)). The sample moves first through the apparatus to a chamber in which\(\ce{H2O}\) is absorbed by a hydrophilic substance and second through a chamber in which \(\ce{CO2}\) is absorbed. The change in weight of each chamber is determined to calculate the weight of \(\ce{H2O}\) and \(\ce{CO2}\). After the masses of \(\ce{H2O}\) and \(\ce{CO2}\) have been determined, they can be used to characterize and calculate the composition of the original sample.
Calculations and determining chemical formulas
Hydrocarbons
Combustion analysis is a standard method of determining a chemical formula of a substance that contains hydrogen and carbon. First, a sample is weighed and then burned in a furnace in the presence of excess oxygen. All of the carbon is converted to carbon dioxide, and the hydrogen is converted to water in this way. Each of these are absorbed in separate compartments, which are weighed before and after the reaction. From these measurements, the chemical formula can be determined.
Generally, the following reaction takes place in combustion analysis:
\[ \ce{C_{a}H_{b} + O2(xs) → aCO2 + b/2 H2O} \nonumber \]
After burning 1.333 g of a hydrocarbon in a combustion analysis apparatus, 1.410 g of \(\ce{H2O}\) and 4.305 g of \(\ce{CO2}\) were produced. Separately, the molar mass of this hydrocarbon was found to be 204.35 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.
Step 1 : Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.
\[ 1.410~\text{g}~\ce{H2O} \times \dfrac{1~\text{mol}~\ce{H2O}}{18.015~\text{g}~\ce{H2O}} \times \dfrac{2~\text{mol H}}{1~\text{mol}~\ce{H2O}} = 0.1565~\text{mol H} \nonumber \]
\[ 4.3051~\text{g}~\ce{CO2} \times \dfrac{1~\text{mol}~\ce{CO2}}{44.010~\text{g}~\ce{CO2}} \times \dfrac{1~\text{mol C}}{1~\text{mol}~\ce{CO2}} = 0.09782 ~\text{mol C} \nonumber \]
Step 2 : Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown
\[ \frac { 0.1565~\mathrm { mol~H } } { 0.09782~\mathrm{ mol~C} } = \frac { 1.600~\mathrm { mol~H } } { 1~\mathrm { mol~C } } = \frac { 16 / 10~\mathrm { mol~H } } { 1~\mathrm { mol~C } } = \frac { 8 / 5~\mathrm { mol~H } } { 1~\mathrm { mol~C } } = \frac { 8~\mathrm { mol~H } } { 5~\mathrm { mol~C } } \nonumber \]
Therefore, the empirical formula is \(\ce{C5H8}\).
Step 3 : To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.
\[ \frac { \text { Molar mass } } { \text { Empirical formula weight } } = \frac { 204.35~\mathrm { g } / \mathrm { mol } } { 68.114~\mathrm { g } / \mathrm { mol } } = 3 \nonumber \]
Therefore, the molecular formula is \(\ce{(C5H8)3}\) or \(\ce{C15H24}\).
After burning 1.082 g of a hydrocarbon in a combustion analysis apparatus, 1.583 g of \(\ce{H2O}\) and 3.315 g of \(\ce{CO2}\) were produced. Separately, the molar mass of this hydrocarbon was found to be 258.52 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.
- Answer
-
The empirical formula is \(\ce{C3H7}\), and the molecular formula is \(\ce{(C3H7)6}\) or\(\ce{ C18H42}\).
Compounds containing carbon, hydrogen, and oxygen
Combustion analysis can also be utilized to determine the empiric and molecular formulas of compounds containing carbon, hydrogen, and oxygen. However, as the reaction is performed in an environment of excess oxygen, the amount of oxygen in the sample can be determined from the sample mass, rather than the combustion data
A 2.0714 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 1.928 g of \(\ce{H2O}\) and 4.709 g of \(\ce{CO2}\) were produced. Separately, the molar mass of the sample was found to be 116.16 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.
Step 1
: Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.
\[ 1.928~\text{g}~\ce{H2O} \times \dfrac{1~\text{mol}~\ce{H2O}}{18.015~\text{g}~\ce{H2O}} \times \dfrac{2~\text{mol H}}{1~\text{mol}~\ce{H2O}} = 0.2140~\text{mol H} \nonumber \]
\[ 4.709~\text{g}~\ce{CO2} \times \dfrac{1~\text{mol}~\ce{CO2}}{44.010~\text{g}~\ce{CO2}} \times \dfrac{1~\text{mol C}}{1~\text{mol}~\ce{CO2}} = 0.1070 ~\text{mol C} \nonumber \]
Step 2 : Using the molar amounts of carbon and hydrogen, calculate the masses of each in the original sample.
\[ 0.2140~\mathrm { mol~H } \times \frac { 1.008~\mathrm { g~H } } { 1~\mathrm { mol~H } } = 0.2157~\mathrm { g~H } \nonumber \]
\[ 0.1070 ~\mathrm{mol~C} \times \frac { 12.011~\mathrm{g~C} } { 1~\mathrm{ mol~C} } = 1.285~\mathrm{g~C} \nonumber \]
Step 3 : Subtract the masses of carbon and hydrogen from the sample mass. Now that the mass of oxygen is known, use this to calculate the molar amount of oxygen in the sample.
\[ 2.0714 \mathrm{ g~sample } - 0.2157~\mathrm{ g~H} - 1.285~\mathrm{ g~C} = 0.5707~\mathrm{ g~O } \nonumber \]
\[ 0.5707 ~\mathrm{mol~O} \times \frac { 1~\mathrm{mol~O} } { 16.00~\mathrm{ g~O} } = 0.03567~\mathrm{g~O} \nonumber \]
Step 4 : Divide each molar amount by the smallest molar amount in order to determine the ratio between the three elements.
\[ \frac { 0.03567~\mathrm { mol~O } } { 0.03567 } = 1.00~\mathrm { mol~O } = 1~\mathrm { mol~O } \nonumber \]
\[ \frac { 0.1070~\mathrm { mol~C } } { 0.03567 } = 3.00 \mathrm { mol~C } = 3~\mathrm { mol~C } \nonumber \]
\[ \frac { 0.2140~\mathrm { mol~H } } { 0.03567 } = 5.999~\mathrm { mol~H } = 6~\mathrm { mol~H } \nonumber \]
Therefore, the empirical formula is \(\ce{C3H6O}\).
Step 5 : To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.
\[ \frac { \text { Molar mass } } { \text { Empirical formula weight } } = \frac { 116.16~\mathrm { g /mol } } { 58.08~\mathrm { g /mol } } = 2 \nonumber \]
Therefore, the molecular formula is \(\ce{(C3H6O)2}\) or \(\ce{C6H12O2}\).
Structure of possible compounds with the molecular formula \(\ce{C6H12O2}\): (a) butylacetate, (b) sec -butyl acetate, (c) tert -butyl acetate, (d) ethyl butyrate, (e) haxanoic acid, (f) isobutyl acetate, (g) methyl pentanoate, and (h) propyl proponoate.
A 4.846 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 4.843 g of \(\ce{H 2 O}\) and 11.83 g of \(\ce{CO 2 }\) were produced. Separately, the molar mass of the sample was found to be 144.22 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.
- Answer
-
The empirical formula is \(\ce{C 4 H 8 O}\), and the molecular formula is (\(\ce{C 4 H 8 O) 2 }\) or \(\ce{C 8 H 16 O 2 }\).
Structure of possible compounds with the molecular formula \(\ce{C8H16O2}\): (a) octanoic acid (caprylic acid), (b) hexyl acetate, (c) pentyl proponate, (d) 2-ethyl hexanoic acid, (e) valproic acid (VPA), (f) cyclohexanedimethanol (CHDM), and (g) 2,2,4,4-tetramethyl-1,3-cyclobutandiol (CBDO).
Binary compounds
By using combustion analysis, the chemical formula of a binary compound containing oxygen can also be determined. This is particularly helpful in the case of combustion of a metal which can result in potential oxides of multiple oxidation states.
A sample of iron weighing 1.7480 g is combusted in the presence of excess oxygen. A metal oxide (\(\ce{Fe_{x}O_{y})}\) is formed with a mass of 2.4982 g. Determine the chemical formula of the oxide product and the oxidation state of Fe.
Step 1 : Subtract the mass of Fe from the mass of the oxide to determine the mass of oxygen in the product.
\[ 2.4982~\mathrm { g~Fe } _ { \mathrm { x } } \mathrm { O } _ { \mathrm { y } } - 1.7480~\mathrm { g~Fe } = 0.7502~\mathrm { g~O } \nonumber \]
Step 2 : Using the molar masses of Fe and O, calculate the molar amounts of each element.
\[ 1.7480 \mathrm { g~Fe } \times \frac { 1 \text { mol Fe } } { 55.845 \text { g Fe } } = 0.031301 \text { mol Fe } \nonumber \]
\[ 0.7502~\text { g } \times \frac { 1 \text { mol O }} { 16.00~\text { g O} } = 0.04689~\text { mol O } \nonumber \]
Step 3 : Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown.
\[ \frac { 0.031301~\text{ mol Fe } } { 0.04689~\mathrm { mol~O } } = \frac { 0.6675~\mathrm { mol~Fe } } { 1~\mathrm { mol~O } } = \frac{ \frac{2}{3} \mathrm { mol~Fe } } { 1~\mathrm { mol~O } } = \frac { 2~\mathrm { mol~Fe } } { 3~\mathrm { mol~O } } \nonumber \]
Therefore, the chemical formula of the oxide is \(\ce{Fe2O3}\), and Fe has a 3+ oxidation state.
A sample of copper weighing 7.295 g is combusted in the presence of excess oxygen. A metal oxide (\(\ce{Cu_{x}O_{y}}\)) is formed with a mass of 8.2131 g. Determine the chemical formula of the oxide product and the oxidation state of Cu.
- Answer
-
The chemical formula is \(\ce{Cu2O}\), and Cu has a 1+ oxidation state..
Bibliography
- J. A. Dumas, Ann. Chem. Pharm. , 1841, 38 , 141.
- H. Goldwhite, J. Chem. Edu. , 1978, 55 , 366.
- A. Lavoisier, Traité Élémentaire de Chimie , 1789, 2 , 493.
- J. Von Liebig, Annalen der Physik und Chemie , 1831, 21 , 1.
- A. Linan and F. A. Williams, Fundamental Aspects of Combustion , Oxford University Press, New York (1993).
- J. M. McBride, "Combustion Analysis," Chemistry 125 , Yale University.
- W. Prout, Philos. T. R. Soc. Lond. , 1827, 117 , 355.
- D. Shriver and P. Atkins, Inorganic Chemistry , 5 th Ed., W. H. Freeman and Co., New York (2009).
- W. Vining et. al., General Chemistry , 1 st Ed., Cengage, Brooks/Cole Cengage Learning , University of Massachusetts Amherst (2014).
- J. Warnatz, U. Maas, and R. W. Dibble, Combustion: Physical and Chemical Fundamentals, Modeling and Simulation, Experiments, Pollutant Formation , 3 rd Ed., Springer, Berlin (2001)


7.3: Applications of Proportions
- solve proportion problems using the five-step method
The Five-Step Method
In [link] we noted that many practical problems can be solved by writing the given information as proportions. Such proportions will be composed of three specified numbers and one unknown number represented by a letter.
The first and most important part of solving a proportion problem is to determine, by careful reading, what the unknown quantity is and to represent it with some letter.
The Five-Step
Method
The five-step method for solving proportion problems:
- By careful reading, determine what the unknown quantity is and represent it with some letter. There will be only one unknown in a problem.
- Identify the three specified numbers.
- Determine which comparisons are to be made and set up the proportion.
- Solve the proportion (using the methods of [link] ).
- Interpret and write a conclusion in a sentence with the appropriate units of measure.
Step 1 is extremely important. Many problems go unsolved because time is not taken to establish what quantity is to be found.
When solving an applied problem, always begin by determining the unknown quantity and representing it with a letter.
Problem Solving
On a map, 2 inches represents 25 miles. How many miles are represented by 8 inches?
Solution
Step 1: The unknown quantity is miles.
Let \(x =\) number of miles represented by 8 inches
Step 2: The three specified numbers are
2 inches
25 miles
8 inches
Step 3: The comparisons are
2 inches to 25 miles \(\to \dfrac{\text{2 inches}}{\text{25 miles}}\)
8 inches to \(x\) miles \(\to \dfrac{\text{8 inches}}{\text{x miles}}\)
Proportions involving ratios and rates are more readily solved by suspending the units while doing the computations. \(\dfrac{2}{25} = \dfrac{8}{x}\)
Step 4: \(\dfrac{2}{25} = \dfrac{8}{x}\) Perform the cross multiplication.
\(\begin{array} {rcl} {2 \cdot x} & = & {8 \cdot 25} \\ {2 \cdot x} & = & {200 \ \ \ \ \ \ \ \text{Divide 200 by 2.}} \\ {x} & = & {\dfrac{200}{2}} \\ {x} & = & {100} \end{array}\)
In step 1, we let \(x\) represent the number of miles. So, \(x\) represents 100 miles.
Step 5: If 2 inches represents 25 miles, then 8 inches represents 100 miles.
An acid solution is composed of 7 parts water to 2 parts acid. How many parts of water are there in a solution composed of 20 parts acid?
Solution
Step 1: The unknown quantity is the number of parts of water.
Let \(x =\) number of parts of water.
Step 2: The three specified numbers are
7 parts water
2 parts acid
20 parts acid
Step 3: The comparisons are
7 parts water to 2 parts acid \(\to \dfrac{7}{2}\)
\(n\) parts water to 20 parts acid \(\to \dfrac{n}{20}\)
\(\dfrac{7}{2} = \dfrac{n}{20}\)
Step 4: \(\dfrac{7}{2} = \dfrac{n}{20}\) Perform the cross multiplication.
\(\begin{array} {rcl} {7 \cdot 20} & = & {2 \cdot n} \\ {140} & = & {2 \cdot n \ \ \ \text{Divide 140 by 2.}} \\ {\dfrac{140}{2}} & = & {n} \\ {70} & = & {n} \end{array}\)
In step 1, we let \(n\) represent the number of parts of water. So, \(n\) represents 70 parts of water.
Step 5: 7 parts water to 2 parts acid indicates 70 parts water to 20 parts acid.
A 5-foot girl casts a \(3 \dfrac{1}{3}\) foot shadow at a particular time of the day. How tall is a person who casts a 3-foot shadow at the same time of the day?
Solution
Step 1: The unknown quantity is the height of the person.
Let \(h = \text{ height of the preson.}\)
Step 2: The three specified numbers are
5 feet ( height of girl)
3\(\dfrac{1}{3}\) feet (length of shadow)
3 feet (length of shadow)
Step 3: The comparisons are
5-foot girl is to \(3 \dfrac{1}{3}\) foot shadow \(\to \dfrac{5}{3 \dfrac{1}{3}}\)
\(h\)-foot person is to 3-foot shadow \(\to \dfrac{h}{3}\)
\(\dfrac{5}{3\dfrac{1}{3}} = \dfrac{h}{3}\)
Step 4:
\(\begin{array} {rcl} {\dfrac{5}{3\dfrac{1}{3}}} & = & {\dfrac{h}{3}} \\ {5 \cdot 3} & = & {3 \dfrac{1}{3} \cdot h} \\ {15} & = & {\dfrac{10}{3} \cdot h \ \ \ \ \text{Divide 15 by } \dfrac{10}{3}} \\ {\dfrac{15}{\dfrac{10}{3}}} & = & {h} \\ {\dfrac{\begin{array} {c} {^3} \\ {\cancel{15}} \end{array}}{1} \cdot \dfrac{3}{\begin{array} {c} {\cancel{10}} \\ {^2} \end{array}}} & = & {h} \\ {\dfrac{9}{2}} & = & {h} \\ {h} & = & {4 \dfrac{1}{2}} \end{array}\)
Step 5: A person who casts a 3-foot shadow at this particular time of the day is \(4 \dfrac{1}{2}\) feet tall.
The ratio of men to women in a particular town is 3 to 5. How many women are there in the town if there are 19,200 men in town?
Solution
Step 1: The unknown quantity is the number of women in town.
Let \(x =\) number of women in town.
Step 2: The three specified numbers are
3
5
19,200
Step 3: The comparisons are 3 men to 5 women \(\to \dfrac{3}{5}\)
19,200 men to \(x\) women \(\to \dfrac{19,200}{x}\)
\(\dfrac{3}{5} = \dfrac{19,200}{x}\)
Step 4:
\(\begin{array} {rcl} {\dfrac{3}{5}} & = & {\dfrac{19,200}{x}} \\ {3 \cdot x} & = & {19,200 \cdot 5} \\ {3 \cdot x} & = & {96,000} \\ {x} & = & {\dfrac{96,000}{3}} \\ {x} & = & {32,000} \end{array}\)
Step 5: There are 32,000 women in town.
The rate of wins to losses of a particular baseball team is \(\dfrac{9}{2}\). How many games did this team lose if they won 63 games?
Solution
Step 1: The unknown quantity is the number of games lost.
Let \(n =\) number of games lost.
Step 2: Since \(\dfrac{9}{2} \to\) means 9 wins to 2 losses, the three specified numbers are
9 (wins)
2 (losses)
63 (wins)
Step 3: The comparisons are
9 wins to 2 losses \(\to \dfrac{9}{2}\)
63 wins to \(n\) losses \(\to \dfrac{63}{n}\)
\(\dfrac{9}{2} = \dfrac{63}{n}\)
Step 4:
\(\begin{array} {rcl} {\dfrac{9}{2}} & = & {\dfrac{63}{n}} \\ {9 \cdot n} & = & {2 \cdot 63} \\ {9 \cdot n} & = & {126} \\ {n} & = & {\dfrac{126}{9}} \\ {n} & = & {14} \end{array}\)
Step 5: This team had 14 losses.
Practice Set A
Solve each problem.
On a map, 3 inches represents 100 miles. How many miles are represented by 15 inches?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
500 miles
Practice Set A
An alcohol solution is composed of 14 parts water to 3 parts alcohol. How many parts of alcohol are in a solution that is composed of 112 parts water?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
24 parts of alcohol
Practice Set A
A \(5 \dfrac{1}{2}\)-foot woman casts a 7-foot shadow at a particular time of the day.
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
\(3 \dfrac{9}{11}\) feet
Practice Set A
The rate of houseplants to outside plants at a nursery is 4 to 9. If there are 384 houseplants in the nursery, how many outside plants are there?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
864 outside plants
Practice Set A
The odds for a particular event occurring are 11 to 2. (For every 11 times the event does occur, it will not occur 2 times.) How many times does the event occur if it does not occur 18 times?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
The event occurs 99 times.
Practice Set A
The rate of passing grades to failing grades in a particular chemistry class is 7272 . If there are 21 passing grades, how many failing grades are there?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
6 failing grades
Exercises
For the following 20 problems, use the five-step method to solve each problem.
Exercise \(\PageIndex{1}\)
On a map, 4 inches represents 50 miles. How many inches represent 300 miles?
- Answer
-
24
Exercise \(\PageIndex{2}\)
On a blueprint for a house, 2 inches represents 3 feet. How many inches represent 10 feet?
Exercise \(\PageIndex{3}\)
A model is built to \(\dfrac{2}{15}\) scale. If a particular part of the model measures 6 inches, how long is the actual structure?
- Answer
-
45 inches
Exercise \(\PageIndex{4}\)
An acid solution is composed of 5 parts acid to 9 parts of water. How many parts of acid are there in a solution that contains 108 parts of water?
Exercise \(\PageIndex{5}\)
An alloy contains 3 parts of nickel to 4 parts of silver. How much nickel is in an alloy that contains 44 parts of silver?
- Answer
-
33 parts
Exercise \(\PageIndex{6}\)
The ratio of water to salt in a test tube is 5 to 2. How much salt is in a test tube that contains 35 ml of water?
Exercise \(\PageIndex{7}\)
The ratio of sulfur to air in a container is \(\dfrac{4}{45}\). How many ml of air are there in a container that contains 207 ml of sulfur?
- Answer
-
2328.75
Exercise \(\PageIndex{8}\)
A 6-foot man casts a 4-foot shadow at a particular time of the day. How tall is a person that casts a 3-foot shadow at that same time of the day?
Exercise \(\PageIndex{9}\)
A \(5\dfrac{1}{2}\)-foot woman casts a \(1\dfrac{1}{2}\)-foot shadow at a particular time of the day. How long a shadow does her \(3\dfrac{1}{2}\)-foot niece cast at the same time of the day?
- Answer
-
\(\dfrac{21}{22}\) feet
Exercise \(\PageIndex{10}\)
A man, who is 6 feet tall, casts a 7-foot shadow at a particular time of the day. How tall is a tree that casts an 84-foot shadow at that same time of the day?
Exercise \(\PageIndex{11}\)
The ratio of books to shelves in a bookstore is 350 to 3. How many books are there in a store that has 105 shelves?
- Answer
-
12,250
Exercise \(\PageIndex{12}\)
The ratio of algebra classes to geometry classes at a particular community college is 13 to 2. How many geometry classes does this college offer if it offers 13 algebra classes?
Exercise \(\PageIndex{13}\)
The odds for a particular event to occur are 16 to 3. If this event occurs 64 times, how many times would you predict it does not occur?
- Answer
-
12
Exercise \(\PageIndex{14}\)
The odds against a particular event occurring are 8 to 3. If this event does occur 64 times, how many times would you predict it does not occur?
Exercise \(\PageIndex{15}\)
The owner of a stationery store knows that a 1-inch stack of paper contains 300 sheets. The owner wishes to stack the paper in units of 550 sheets. How many inches tall should each stack be?
- Answer
-
\(1\dfrac{5}{6}\)
Exercise \(\PageIndex{16}\)
A recipe that requires 6 cups of sugar for 15 servings is to be used to make 45 servings. How much sugar will be needed?
Exercise \(\PageIndex{17}\)
A pond loses \(7\dfrac{1}{2}\) gallons of water every 2 days due to evaporation. How many gallons of water are lost, due to evaporation, in \(\dfrac{1}{2}\) day?
- Answer
-
\(1\dfrac{7}{8}\)
Exercise \(\PageIndex{18}\)
A photograph that measures 3 inches wide and \(4\dfrac{1}{2}\) inches high is to be enlarged so that it is 5 inches wide. How high will it be?
Exercise \(\PageIndex{19}\)
If 25 pounds of fertilizer covers 400 square feet of grass, how many pounds will it take to cover 500 square feet of grass?
- Answer
-
\(31 \dfrac{1}{4}\)
Exercise \(\PageIndex{20}\)
Every \(\dfrac{1}{2}\) teaspoons of a particular multiple vitamin, in granular form, contains 0.65 the minimum daily requirement of vitamin C. How many teaspoons of this vitamin are required to supply 1.25 the minimum daily requirement?
Exercises for Review
Exercise \(\PageIndex{21}\)
Find the product, \(818 \cdot 0\).
- Answer
-
0
Exercise \(\PageIndex{22}\)
Determine the missing numerator: \(\dfrac{8}{15} = \dfrac{N}{90}\).
Exercise \(\PageIndex{23}\)
Find the value of \(\dfrac{\dfrac{3}{10} + \dfrac{4}{12}}{\dfrac{19}{20}}\).
- Answer
-
\(\dfrac{2}{3}\)
Exercise \(\PageIndex{24}\)
Subtract 0.249 from the sum of 0.344 and 0.612.
Exercise \(\PageIndex{25}\)
Solve the proportion: \(\dfrac{6}{x} = \dfrac{36}{30}\).
- Answer
-
5


8.1: Polar Coordinates and Polar Graphs
-
- Last updated
- Save as PDF
Outcomes
- Understand polar coordinates.
- Convert points between Cartesian and polar coordinates.
You have likely encountered the Cartesian coordinate system in many aspects of mathematics. There is an alternative way to represent points in space, called polar coordinates . The idea is suggested in the following picture.
Consider the point above, which would be specified as \((x,y)\) in Cartesian coordinates. We can also specify this point using polar coordinates, which we write as \(\left( r, \theta \right)\). The number \(r\) is the distance from the origin\(\left( 0,0\right)\) to the point, while \(\theta\) is the angle shown between the positive \(x\) axis and the line from the origin to the point. In this way, the point can be specified in polar coordinates as \(\left(r, \theta \right)\).
Now suppose we are given an ordered pair \(\left( r,\theta \right)\) where \(r\) and \(\theta\) are real numbers. We want to determine the point specified by this ordered pair. We can use \(\theta\) to identify a ray from the origin as follows. Let the ray pass from \(\left( 0,0\right)\) through the point \(\left( \cos \theta ,\sin \theta \right)\) as shown.
The ray is identified on the graph as the line from the origin, through the point \((\mbox{cos}(\theta),\mbox{sin}(\theta))\). Now if \(r>0,\) go a distance equal to \(r\) in the direction of the displayed arrow starting at \((0,0)\). If \(r<0,\) move in the opposite direction a distance of \(\left\vert r\right\vert\). This is the point determined by \(\left( r,\theta \right)\).
It is common to assume that \(\theta\) is in the interval \([0,2\pi )\) and \(r>0.\) In this case, there is a very simple relationship between the Cartesian and polar coordinates, given by \[x=r\cos \left( \theta \right) ,\ \ y=r\sin \left( \theta \right) \label{cartpolcoord}\]
These equations demonstrate how to find the Cartesian coordinates when we are given the polar coordinates of a point. They can also be used to find the polar coordinates when we know \(\left( x, y \right)\). A simpler way to do this is the following equations: \[\begin{array}{l} r = \sqrt{x^2 + y^2} \\ \\ \mbox{tan} \left(\theta\right) = \frac{y}{x} \end{array} \label{polcartcoord}\]
In the next example, we look at how to find the Cartesian coordinates of a point specified by polar coordinates.
Example \(\PageIndex{1}\): Finding Cartesian Coordinates
The polar coordinates of a point in the plane are \(\left( 5,\pi /6\right)\). Find the Cartesian coordinates of this point.
Solution
The point is specified by the polar coordinates \(\left( 5,\pi /6\right)\). Therefore \(r=5\) and \(\theta = \pi /6\). From \(\eqref{cartpolcoord}\) \[x= r \cos \left( \theta \right)= 5\cos \left( \frac{\pi }{6}\right) = \frac{5}{2}\sqrt{3}\nonumber \] \[y= r \sin \left( \theta \right) = 5\sin \left( \frac{\pi }{6}\right) = \frac{5}{2}\nonumber \] Thus the Cartesian coordinates are \(\left( \frac{5}{2}\sqrt{3}, \frac{5}{2}\right)\). The point is shown in the below graph.
Consider the following example of the case where \(r < 0\).
Example \(\PageIndex{2}\): Finding Cartesian Coordinates
The polar coordinates of a point in the plane are \(\left( -5,\pi /6\right) .\) Find the Cartesian coordinates.
Solution
For the point specified by the polar coordinates \(\left( -5, \pi /6 \right)\), \(r=-5\), and \(x\theta = \pi /6\). From \(\eqref{cartpolcoord}\) \[x= r \cos \left( \theta \right)= -5\cos \left( \frac{\pi }{6}\right) = -\frac{5}{2}\sqrt{3}\nonumber \] \[y= r \sin \left( \theta \right) = -5\sin \left( \frac{\pi }{6}\right) = -\frac{5}{2}\nonumber \] Thus the Cartesian coordinates are \(\left( -\frac{5}{2}\sqrt{3}, -\frac{5}{2}\right)\). The point is shown in the following graph.
Recall from the previous example that for the point specified by \(\left( 5, \pi /6 \right)\), the Cartesian coordinates are \(\left( \frac{5}{2}\sqrt{3}, \frac{5}{2}\right)\). Notice that in this example, by multiplying \(r\) by \(-1\), the resulting Cartesian coordinates are also multiplied by \(-1\).
The following picture exhibits both points in the above two examples to emphasize how they are just on opposite sides of \(\left( 0,0\right)\) but at the same distance from \(\left( 0,0\right)\).
In the next two examples, we look at how to convert Cartesian coordinates to polar coordinates.
Example \(\PageIndex{3}\): Finding Polar Coordinates
Suppose the Cartesian coordinates of a point are \(\left( 3,4\right)\). Find a pair of polar coordinates which correspond to this point.
Solution
Using equation \(\eqref{polcartcoord}\), we can find \(r\) and \(\theta\). Hence \(r=\sqrt{3^{2}+4^{2}}=5\). It remains to identify the angle \(\theta\) between the positive \(x\) axis and the line from the origin to the point. Since both the \(x\) and \(y\) values are positive, the point is in the first quadrant. Therefore, \(\theta\) is between \(0\) and \(\pi /2\ \). Using this and \(\eqref{polcartcoord}\), we have to solve: \[\mbox{tan}\left(\theta \right)=\frac{4}{3}\nonumber \] Conversely, we can use equation \(\eqref{cartpolcoord}\) as follows: \[3=5\cos \left( \theta \right)\nonumber \] \[4 = 5\sin \left( \theta \right)\nonumber \] Solving these equations, we find that, approximately, \(\theta =0.\, 927\,295\) radians.
Consider the following example.
Example \(\PageIndex{4}\): Finding Polar Coordinates
Suppose the Cartesian coordinates of a point are \(\left( -\sqrt{3},1\right)\). Find the polar coordinates which correspond to this point.
Solution
Given the point \(\left( -\sqrt{3}, 1\right)\), \[\begin{aligned} r &= \sqrt{ 1^2 + (-\sqrt{3})^2}\\ &= \sqrt{1 + 3}\\ &=2\end{aligned}\] In this case, the point is in the second quadrant since the \(x\) value is negative and the \(y\) value is positive. Therefore, \(\theta\) will be between \(\pi/2\) and \(\pi\). Solving the equations \[-\sqrt{3}= 2 \cos \left(\theta\right)\nonumber \] \[1 = 2 \sin \left( \theta\right)\nonumber \] we find that \(\theta = 5\pi /6.\) Hence the polar coordinates for this point are \(\left(2, 5\pi /6 \right)\).
Consider this example. Suppose we used \(r=-2\) and \(\theta =2\pi -\left( \pi /6\right) = 11\pi /6\). These coordinates specify the same point as above. Observe that there are infinitely many ways to identify this particular point with polar coordinates. In fact, every point can be represented with polar coordinates in infinitely many ways. Because of this, it will usually be the case that \(\theta\) is confined to lie in some interval of length \(2\pi\) and \(r>0\), for real numbers \(r\) and \(\theta\).
Just as with Cartesian coordinates, it is possible to use relations between the polar coordinates to specify points in the plane. The process of sketching the graphs of these relations is very similar to that used to sketch graphs of functions in Cartesian coordinates. Consider a relation between polar coordinates of the form, \(r=f\left( \theta \right)\). To graph such a relation, first make a table of the form
| \(\theta\) | \(r\) |
|---|---|
| \(\theta_1\) | \(f(\theta_1)\) |
| \(\theta_2\) | \(f(\theta_2)\) |
| \(\vdots\) | \(\vdots\) |
Graph the resulting points and connect them with a curve. The following picture illustrates how to begin this process.
To find the point in the plane corresponding to the ordered pair \(\left( f\left( \theta \right) ,\theta \right)\), we follow the same process as when finding the point corresponding to \(\left( r, \theta \right)\).
Consider the following example of this procedure, incorporating computer software.
Example \(\PageIndex{5}\)
Graphing a Polar Equation Graph the polar equation \(r=1+\cos \theta\).
Solution
We will use the computer software Maple to complete this example. The command which produces the polar graph of the above equation is: \(>\) plot(1+cos(t),t= 0..2*Pi,coords=polar). Here we use \(t\) to represent the variable \(\theta\) for convenience. The command tells Maple that \(r\) is given by \(1+\cos \left( t\right)\) and that \(t\in \left[ 0,2\pi \right]\).
The above graph makes sense when considered in terms of trigonometric functions. Suppose \(\theta =0,r=2\) and let \(\theta\) increase to \(\pi /2\). As \(\theta\) increases, \(\cos \theta\) decreases to 0. Thus the line from the origin to the point on the curve should get shorter as \(\theta\) goes from \(0\) to \(\pi /2\). As \(\theta\) goes from \(\pi /2\) to \(\pi\), \(\cos \theta\) decreases, eventually equaling \(-1\) at \(\theta =\pi\). Thus \(r=0\) at this point. This scenario is depicted in the above graph, which shows a function called a cardioid .
The following picture illustrates the above procedure for obtaining the polar graph of \(r=1+cos(\theta)\). In this picture, the concentric circles correspond to values of \(r\) while the rays from the origin correspond to the angles which are shown on the picture. The dot on the ray corresponding to the angle \(\pi/6\) is located at a distance of \(r = 1+cos(\pi/6)\) from the origin. The dot on the ray corresponding to the angle \(\pi/3\) is located at a distance of \(r = 1+cos(\pi/3)\) from the origin and so forth. The polar graph is obtained by connecting such points with a smooth curve, with the result being the figure shown above.
Consider another example of constructing a polar graph.
Example \(\PageIndex{6}\): A Polar Graph
Graph \(r=1+2\cos \theta\) for \(\theta \in \left[ 0,2\pi \right]\).
Solution
The graph of the polar equation \(r=1+2\cos \theta\) for \(\theta \in \left[ 0,2\pi \right]\) is given as follows.
To see the way this is graphed, consider the following picture. First the indicated points were graphed and then the curve was drawn to connect the points. When done by a computer, many more points are used to create a more accurate picture.
Consider first the following table of points.
| \(\theta\) | \(\pi /6\) | \(\pi /3\) | \(\pi /2\) | \(5\pi /6\) | \(\pi\) | \(4\pi /3\) | \(7\pi /6\) | \(5\pi /3\) |
|---|---|---|---|---|---|---|---|---|
| \(r\) | \(\sqrt{3}+1\) | \(2\) | \(1\) | \(1-\sqrt{3}\) | \(-1\) | \(0\) | \(1-\sqrt{3}\) | \(2\) |
Note how some entries in the table have \(r<0.\) To graph these points, simply move in the opposite direction. These types of points are responsible for the small loop on the inside of the larger loop in the graph.
The process of constructing these graphs can be greatly facilitated by computer software. However, the use of such software should not replace understanding the steps involved.
The next example shows the graph for the equation \(r=3+\sin \left( \displaystyle \frac{7\theta }{6}\right)\). For complicated polar graphs, computer software is used to facilitate the process.
Example \(\PageIndex{7}\): A Polar Graph
Graph \(r=3+\sin \left( \displaystyle \frac{7\theta }{6} \right)\) for \(\theta \in \left[ 0,14\pi \right]\).
Solution
The next example shows another situation in which \(r\) can be negative.
Example \(\PageIndex{8}\): A Polar Graph: Negative \(r\)
Graph \(r=3sin(4\theta)\) for \(\theta \in \left[ 0,2\pi \right]\).
Solution
We conclude this section with an interesting graph of a simple polar equation.
Example \(\PageIndex{9}\): The Graph of a Spiral
Graph \(r=\theta\) for \(\theta \in [0,2\pi]\).
Solution
The graph of this polar equation is a spiral. This is the case because as \(\theta\) increases, so does \(r\).
In the next section, we will look at two ways of generalizing polar coordinates to three dimensions.


13.1.4: The Planetarium
The Planetarium
The planetarium as we know it today is essentially a new technology. Often called the planetarium theater or dome, the room itself is marked by a domed ceiling, onto which astronomical bodies are projected.
The Greeks are credited with creating a device which could predict the motions of the Sun, Moon, and the planets known at the time (Mercury, Venus, Mars, Jupiter, and Saturn). These devices became known as an orrery. Other early planetarium-like models included the painting of stars and constellations on walls and ceilings, and large domes with theater-like star effects.
The first modern-era planetarium was designed and built by the Carl Zeiss Company in Jena, Germany and installed in the Deutsches Museum in Munich. The stars were projected onto a plaster dome.
Advances in electronics and Space Race demand for science education pushed more museums, colleges and school districts to install planetariums. Over the years, the planetarium has developed into a multi-dimensional theater, combining all-dome film systems, special effects, and even all-dome planetarium projectors to give students and audiences a modern look at astronomy and the sciences.


22.7: Equivalent Circuit Model for Reception
-
- Last updated
- Save as PDF
In this section, we begin to address antennas as devices that convert incident electromagnetic waves into potentials and currents in a circuit. It is convenient to represent this process in the form of a Thévenin equivalent circuit. The particular circuit addressed in this section is shown in Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): Thévenin equivalent circuit for an antenna in the presence of an incident electromagnetic wave. (CC BY-SA 4.0 (modified); S. Lally)The circuit consists of a voltage source \(\widetilde{V}_{OC}\) and a series impedance \(Z_A\). The source potential \(\widetilde{V}_{OC}\) is the potential at the terminals of the antenna when there is no termination; i.e., when the antenna is open-circuited. The series impedance \(Z_A\) is the output impedance of the circuit, and so determines the magnitude and phase of the current at the terminals once a load is connected. Given \(\widetilde{V}_{OC}\) and the current through the equivalent circuit, it is possible to determine the power delivered to the load. Thus, this model is quite useful, but only if we are able to determine \(\widetilde{V}_{OC}\) and \(Z_A\). This section provides an informal derivation of these quantities that is sufficient to productively address the subsequent important topics of effective aperture and impedance matching of receive antennas. 1
Vector effective length
With no derivation required, we can deduce the following about \(\widetilde{V}_{OC}\):
- \(\widetilde{V}_{OC}\) must depend on the incident electric field intensity \(\widetilde{\bf E}^i\). Presumably the relationship is linear, so \(\widetilde{V}_{OC}\) is proportional to the magnitude of \(\widetilde{\bf E}^i\).
-
Since \(\widetilde{\bf E}^i\) is a vector whereas \(\widetilde{V}_{OC}\) is a scalar, there must be some vector \({\bf l}_e\) for which
\[\widetilde{V}_{OC}=\widetilde{\bf E}^i\cdot{\bf l}_e \label{m0206_Voc} \]
- Since \(\widetilde{\bf E}^i\) has SI base units of V/m and \(\widetilde{V}_{OC}\) has SI base units of V, \({\bf l}_e\) must have SI base units of m; i.e., length.
- We expect that \(\widetilde{V}_{OC}\) increases as the size of the antenna increases, so the magnitude of \({\bf l}_e\) likely increases with the size of the antenna.
- The direction of \({\bf l}_e\) must be related to the orientation of the incident electric field relative to that of the antenna, since this is clearly important yet we have not already accounted for this.
It may seem at this point that \({\bf l}_e\) is unambiguously determined, and we need merely to derive its value. However, this is not the case. There are in fact multiple unique definitions of \({\bf l}_e\) that will reduce the vector \(\widetilde{\bf E}^i\) to the observed scalar \(\widetilde{V}_{OC}\) via Equation \ref{m0206_Voc}. In this section, we shall employ the most commonly-used definition, in which \({\bf l}_e\) is referred to as vector effective length . Following this definition, the scalar part \(l_e\) of \({\bf l}_e=\hat{\bf l} l_e\) is commonly referred to as any of the following: effective length (the term used in this book), effective height , or antenna factor .
In this section, we shall merely define vector effective length, and defer a formal derivation to Section 10.11. In this definition, we arbitrarily set \(\hat{\bf l}\), the real-valued unit vector indicating the direction of \({\bf l}_e\), equal to the direction in which the electric field transmitted from this antenna would be polarized in the far field. For example, consider a \(\hat{\bf z}\)-oriented electrically-short dipole (ESD) located at the origin. The electric field transmitted from this antenna would have only a \(\hat{\bf \theta}\) component, and no \(\hat{\bf \phi}\) component (and certainly no \(\hat{\bf r}\) component). Thus, \(\hat{\bf l}=\hat{\bf \theta}\) in this case.
Applying this definition, \(\widetilde{\bf E}^i \cdot \hat{\bf l}\) yields the scalar component of \(\widetilde{\bf E}^i\) that is co-polarized with electric field radiated by the antenna when transmitting. Now \(l_e\) is uniquely defined to be the factor that converts this component into \(\widetilde{V}_{OC}\). Summarizing:
The vector effective length \({\bf l}_e = \hat{\bf l} l_e\) is defined as follows: \(\hat{\bf l}\) is the real-valued unit vector corresponding to the polarization of the electric field that would be transmitted from the antenna in the far field. Subsequently, the effective length \(l_e\) is
\[l_e \triangleq \frac{\widetilde{V}_{OC}}{\widetilde{\bf E}^i \cdot \hat{\bf l}} \nonumber \]
where \(\widetilde{V}_{OC}\) is the open-circuit potential induced at the antenna terminals in response to the incident electric field intensity \(\widetilde{\bf E}^i\).
While this definition yields an unambiguous value for \(l_e\), it is not yet clear what that value is. For most antennas, effective length is quite difficult to determine directly, and one must instead determine effective length indirectly from the transmit characteristics via reciprocity. This approach is relatively easy (although still quite a bit of effort) for thin dipoles, and is presented in Section 10.11.
To provide an example of how effective length works right away, consider the \(\hat{\bf z}\)-oriented ESD described earlier in this section. Let the length of this ESD be \(L\). Let \(\widetilde{\bf E}^i\) be a \(\hat{\bf \theta}\)-polarized plane wave arriving at the ESD. The ESD is open-circuited, so the potential induced in its terminals is \(\widetilde{V}_{OC}\). One observes the following:
- When \(\widetilde{\bf E}^i\) arrives from anywhere in the \(\theta=\pi/2\) plane (i.e., broadside to the ESD), \(\widetilde{\bf E}^i\) points in the \(-\hat{\bf z}\) direction, and we find that \(l_e\approx L/2\). It should not be surprising that \(l_e\) is proportional to \(L\); this expectation was noted earlier in this section.
- When \(\widetilde{\bf E}^i\) arrives from the directions \(\theta=0\) or \(\theta=\pi\) – i.e., along the axis of the ESD – \(\widetilde{\bf E}^i\) is perpendicular to the axis of the ESD. In this case, we find that \(l_e\) equals zero.
Taken together, these findings suggest that \(l_e\) should contain a factor of \(\sin\theta\). We conclude that the vector effective length for a \(\hat{\bf z}\)-directed ESD of length \(L\) is
\[{\bf l}_e \approx \hat{\bf \theta}\frac{L}{2}\sin\theta ~~~~~ \mbox{(ESD)} \label{m0206_eVEL-ESD} \]
Example \(\PageIndex{1}\): Potential induced in an ESD
A thin straight dipole of length 10 cm is located at the origin and aligned with the \(z\)-axis. A plane wave is incident on the dipole from the direction \((\theta=\pi/4,\phi=\pi/2)\). The frequency of the wave is 30 MHz. The magnitude of the incident electric field is \(10~\mu\)V/m (rms). What is the magnitude of the induced open-circuit potential when the electric field is (a) \(\hat{\bf \theta}\)-polarized and (b) \(\hat{\bf \phi}\)-polarized?
Solution
The wavelength in this example is \(c/f\cong 10\) m, so this dipole is electrically-short. Using Equation \ref{m0206_eVEL-ESD}:
\[\begin{aligned} {\bf l}_e & \approx \hat{\bf \theta}\frac{10~\mbox{cm}}{2}\sin\frac{\pi}{4} \nonumber \\ & \approx \hat{\bf \theta}\left(3.54~\mbox{cm}\right) \nonumber\end{aligned} \nonumber \]
Thus, the effective length \(l_e=3.54\) cm. When the electric field is \(\hat{\bf \theta}\)-polarized, the magnitude of the induced open-circuit voltage is
\[\begin{aligned} \left|\widetilde{V}_{OC}\right| &=\left| \widetilde{\bf E}^i \cdot {\bf l}_e \right| \nonumber \\ &\approx \left(10~\mu\mbox{V/m}\right)\hat{\bf \theta} \cdot \hat{\bf \theta}\left(3.54~\mbox{cm}\right) \nonumber \\ &\approx \underline{354~\mbox{nV}~~\mbox{rms}}~~~ \mbox{(a)} \nonumber\end{aligned} \nonumber \]
When the electric field is \(\hat{\bf \phi}\)-polarized:
\[\begin{aligned} \left|\widetilde{V}_{OC}\right| &\approx \left(10~\mu\mbox{V/m}\right)\hat{\bf \phi} \cdot \hat{\bf \theta}\left(3.54~\mbox{cm}\right) \nonumber \\ &\approx \underline{0} ~~~ \mbox{(b)} \nonumber\end{aligned} \nonumber \]
This is because the polarization of the incident electric field is orthogonal to that of the ESD. In fact, the answer to part (b) is zero for any angle of incidence \((\theta,\phi)\).
Output impedance
The output impedance \(Z_A\) is somewhat more difficult to determine without a formal derivation, which is presented in Section 10.12. For the purposes of this section, it suffices to jump directly to the result:
The output impedance \(Z_A\) of the equivalent circuit for an antenna in the receive case is equal to the input impedance of the same antenna in the transmit case.
This remarkable fact is a consequence of the reciprocity property of antenna systems, and greatly simplifies the analysis of receive antennas.
Now a demonstration of how the antenna equivalent circuit can be used to determine the power delivered by an antenna to an attached electrical circuit:
Example \(\PageIndex{2}\): Power captured by an ESD
Continuing with part (a) of Example \(\PageIndex{1}\): If this antenna is terminated into a conjugate-matched load, then what is the power delivered to that load? Assume the antenna is lossless.
Solution
First, we determine the impedance \(Z_A\) of the equivalent circuit of the antenna. This is equal to the input impedance of the antenna in transmission. Let \(R_A\) and \(X_A\) be the real and imaginary parts of this impedance; i.e., \(Z_A=R_A+jX_A\). Further, \(R_A\) is the sum of the radiation resistance \(R_{rad}\) and the loss resistance. The loss resistance is zero because the antenna is lossless. Since this is an ESD:
\[R_{rad} \approx 20\pi^2 \left(\frac{L}{\lambda}\right)^2 \nonumber \]
Therefore, \(R_A = R_{rad}\approx 4.93\) m\(\Omega\). We do not need to calculate \(X_A\), as will become apparent in the next step.
A conjugate-matched load has impedance \(Z_A^*\), so the potential \(\widetilde{V}_L\) across the load is
\[\widetilde{V}_L = \widetilde{V}_{OC}\frac{Z_A^*}{Z_A+Z_A^*} = \widetilde{V}_{OC}\frac{Z_A^*}{2R_A} \nonumber \]
The current \(\widetilde{I}_L\) through the load is
\[\widetilde{I}_L = \frac{\widetilde{V}_{OC}}{Z_A+Z_A^*} = \frac{\widetilde{V}_{OC}}{2R_A} \nonumber \]
Taking \(\widetilde{V}_{OC}\) as an RMS quantity, the power \(P_L\) delivered to the load is
\[P_L = \mbox{Re}\left\{V_L I_L^*\right\} = \frac{\left|\widetilde{V}_{OC}\right|^2}{4R_A} \nonumber \]
In part (a) of Example \(\PageIndex{1}\), \(\left|\widetilde{V}_{OC}\right|\) is found to be \(\approx 354\) nV rms, so \(P_L \approx \underline{6.33 \text{ pW}}\).
- Formal derivations of these quantities are provided in subsequent sections. The starting point is the section on reciprocity.↩


12.9: Glossary- The Digestive System
absorption passage of digested products from the intestinal lumen through mucosal cells and into the bloodstream or lacteals
accessory digestive organ includes teeth, tongue, salivary glands, gallbladder, liver, and pancreas
accessory duct (also, duct of Santorini) duct that runs from the pancreas into the duodenum
acinus cluster of glandular epithelial cells in the pancreas that secretes pancreatic juice in the pancreas
alimentary canal continuous muscular digestive tube that extends from the mouth to the anus
anal canal final segment of the large intestine
anal column long fold of mucosa in the anal canal
anal sinus between anal columns
appendix (vermiform appendix) coiled tube attached to the cecum
ascending colon first region of the colon
bacterial flora bacteria in the large intestine
bile alkaline solution produced by the liver and important for the emulsification of lipids
bile canaliculus small duct between hepatocytes that collects bile
bilirubin main bile pigment, which is responsible for the brown color of feces
body mid-portion of the stomach
bolus mass of chewed food
brush border fuzzy appearance of the small intestinal mucosa created by microvilli
cardia (also, cardiac region) part of the stomach surrounding the cardiac orifice (esophageal hiatus)
cecum pouch forming the beginning of the large intestine
cementum bone-like tissue covering the root of a tooth
central vein vein that receives blood from hepatic sinusoids
cephalic phase (also, reflex phase) initial phase of gastric secretion that occurs before food enters the stomach
chemical digestion enzymatic breakdown of food
chief cell gastric gland cell that secretes pepsinogen
chyme soupy liquid created when food is mixed with digestive juices
circular fold (also, plica circulare) deep fold in the mucosa and submucosa of the small intestine
colon part of the large intestine between the cecum and the rectum
common bile duct structure formed by the union of the common hepatic duct and the gallbladder’s cystic duct
common hepatic duct duct formed by the merger of the two hepatic ducts
crown portion of tooth visible superior to the gum line
cuspid (also, canine) pointed tooth used for tearing and shredding food
cystic duct duct through which bile drains and enters the gallbladder
deciduous tooth one of 20 “baby teeth”
defecation elimination of undigested substances from the body in the form of feces
deglutition three-stage process of swallowing
dens tooth
dentin bone-like tissue immediately deep to the enamel of the crown or cementum of the root of a tooth
dentition set of teeth
descending colon part of the colon between the transverse colon and the sigmoid colon
duodenal gland (also, Brunner’s gland) mucous-secreting gland in the duodenal submucosa
duodenum first part of the small intestine, which starts at the pyloric sphincter and ends at the jejunum
enamel covering of the dentin of the crown of a tooth
enteroendocrine cell gastric gland cell that releases hormones
enterohepatic circulation recycling mechanism that conserves bile salts
enteropeptidase intestinal brush-border enzyme that activates trypsinogen to trypsin
epiploic appendage small sac of fat-filled visceral peritoneum attached to teniae coli
esophagus muscular tube that runs from the pharynx to the stomach
external anal sphincter voluntary skeletal muscle sphincter in the anal canal
fauces opening between the oral cavity and the oropharynx
feces semisolid waste product of digestion
flatus gas in the intestine
fundus dome-shaped region of the stomach above and to the left of the cardia
G cell gastrin-secreting enteroendocrine cell
gallbladder accessory digestive organ that stores and concentrates bile
gastric emptying process by which mixing waves gradually cause the release of chyme into the duodenum
gastric gland gland in the stomach mucosal epithelium that produces gastric juice
gastric phase phase of gastric secretion that begins when food enters the stomach
gastric pit narrow channel formed by the epithelial lining of the stomach mucosa
gastrin peptide hormone that stimulates secretion of hydrochloric acid and gut motility
gastrocolic reflex propulsive movement in the colon activated by the presence of food in the stomach
gastroileal reflex long reflex that increases the strength of segmentation in the ileum
gingiva gum
haustrum small pouch in the colon created by tonic contractions of teniae coli
haustral contraction slow segmentation in the large intestine
hepatic artery artery that supplies oxygenated blood to the liver
hepatic lobule hexagonal-shaped structure composed of hepatocytes that radiate outward from a central vein
hepatic portal vein vein that supplies deoxygenated nutrient-rich blood to the liver
hepatic sinusoid blood capillaries between rows of hepatocytes that receive blood from the hepatic portal vein and the branches of the hepatic artery
hepatic vein vein that drains into the inferior vena cava
hepatocytes major functional cells of the liver
hepatopancreatic ampulla (also, ampulla of Vater) bulb-like point in the wall of the duodenum where the bile duct and main pancreatic duct unite
hepatopancreatic sphincter (also, sphincter of Oddi) sphincter regulating the flow of bile and pancreatic juice into the duodenum
hydrochloric acid (HCl) digestive acid secreted by parietal cells in the stomach
ileocecal sphincter sphincter located where the small intestine joins with the large intestine
ileum end of the small intestine between the jejunum and the large intestine
incisor midline, chisel-shaped tooth used for cutting into food
ingestion taking food into the GI tract through the mouth
internal anal sphincter involuntary smooth muscle sphincter in the anal canal
intestinal gland (also, crypt of Lieberkühn) gland in the small intestinal mucosa that secretes intestinal juice
intestinal juice mixture of water and mucus that helps absorb nutrients from chyme
intrinsic factor glycoprotein required for vitamin B 12 absorption in the small intestine
intestinal phase phase of gastric secretion that begins when chyme enters the intestine
jejunum middle part of the small intestine between the duodenum and the ileum
labium lip
labial frenulum midline mucous membrane fold that attaches the inner surface of the lips to the gums
lacteal lymphatic capillary in the villi
large intestine terminal portion of the alimentary canal
laryngopharynx part of the pharynx that functions in respiration and digestion
left colic flexure (also, splenic flexure) point where the transverse colon curves below the inferior end of the spleen
lingual frenulum mucous membrane fold that attaches the bottom of the tongue to the floor of the mouth
lingual lipase digestive enzyme from glands in the tongue that acts on triglycerides
liver largest gland in the body whose main digestive function is the production of bile
lower esophageal sphincter smooth muscle sphincter that regulates food movement from the esophagus to the stomach
main pancreatic duct (also, duct of Wirsung) duct through which pancreatic juice drains from the pancreasmajor duodenal papilla point at which the hepatopancreatic ampulla opens into the duodenum
mass movement long, slow, peristaltic wave in the large intestine
mastication chewing
mechanical digestion chewing, mixing, and segmentation that prepares food for chemical digestion
mesoappendix mesentery of the appendix
microvillus small projection of the plasma membrane of the absorptive cells of the small intestinal mucosa
migrating motility complex form of peristalsis in the small intestine
mixing wave unique type of peristalsis that occurs in the stomach
molar tooth used for crushing and grinding food
motilin hormone that initiates migrating motility complexes
motility movement of food through the GI tract
mucosa innermost lining of the alimentary canal
mucosal barrier protective barrier that prevents gastric juice from destroying the stomach itself
mucous neck cell gastric gland cell that secretes a uniquely acidic mucus
muscularis muscle (skeletal or smooth) layer of the alimentary canal wall
myenteric plexus (plexus of Auerbach) major nerve supply to alimentary canal wall; controls motility
oral cavity (also, buccal cavity) mouth
oral vestibule part of the mouth bounded externally by the cheeks and lips, and internally by the gums and teeth
oropharynx part of the pharynx continuous with the oral cavity that functions in respiration and digestion
palatoglossal arch muscular fold that extends from the lateral side of the soft palate to the base of the tongue
palatopharyngeal arch muscular fold that extends from the lateral side of the soft palate to the side of the pharynx
pancreas accessory digestive organ that secretes pancreatic juice
pancreatic juice secretion of the pancreas containing digestive enzymes and bicarbonate
parietal cell gastric gland cell that secretes hydrochloric acid and intrinsic factor
parotid gland one of a pair of major salivary glands located inferior and anterior to the ears
pectinate line horizontal line that runs like a ring, perpendicular to the inferior margins of the anal sinuses
pepsinogen inactive form of pepsin
peristalsis muscular contractions and relaxations that propel food through the GI tract
permanent tooth one of 32 adult teeth
pharynx throat
porta hepatis “gateway to the liver” where the hepatic artery and hepatic portal vein enter the liver
portal triad bile duct, hepatic artery branch, and hepatic portal vein branch
premolar (also, bicuspid) transitional tooth used for mastication, crushing, and grinding food
propulsion voluntary process of swallowing and the involuntary process of peristalsis that moves food through the digestive tract
pulp cavity deepest portion of a tooth, containing nerve endings and blood vessels
pyloric antrum wider, more superior part of the pylorus
pyloric canal narrow, more inferior part of the pylorus
pyloric sphincter sphincter that controls stomach emptying
pylorus lower, funnel-shaped part of the stomach that is continuous with the duodenum
rectal valve one of three transverse folds in the rectum where feces is separated from flatus
rectum part of the large intestine between the sigmoid colon and anal canal
reticuloendothelial cell (also, Kupffer cell) phagocyte in hepatic sinusoids that filters out material from venous blood from the alimentary canal
retroperitoneal located posterior to the peritoneum
right colic flexure (also, hepatic flexure) point, at the inferior surface of the liver, where the ascending colon turns abruptly to the left
root portion of a tooth embedded in the alveolar processes beneath the gum line
ruga fold of alimentary canal mucosa and submucosa in the empty stomach and other organs
saccharolytic fermentation anaerobic decomposition of carbohydrates
saliva aqueous solution of proteins and ions secreted into the mouth by the salivary glands
salivary amylase digestive enzyme in saliva that acts on starch
salivary gland an exocrine gland that secretes a digestive fluid called saliva
salivation secretion of saliva
segmentation alternating contractions and relaxations of non-adjacent segments of the intestine that move food forward and backward, breaking it apart and mixing it with digestive juices
serosa outermost layer of the alimentary canal wall present in regions within the abdominal cavity
sigmoid colon end portion of the colon, which terminates at the rectum
small intestine section of the alimentary canal where most digestion and absorption occurs
soft palate posterior region of the bottom portion of the nasal cavity that consists of skeletal muscle
stomach alimentary canal organ that contributes to chemical and mechanical digestion of food from the esophagus before releasing it, as chyme, to the small intestine
sublingual gland one of a pair of major salivary glands located beneath the tongue
submandibular gland one of a pair of major salivary glands located in the floor of the mouth
submucosa layer of dense connective tissue in the alimentary canal wall that binds the overlying mucosa to the underlying muscularis
submucosal plexus (plexus of Meissner) nerve supply that regulates activity of glands and smooth muscle
tenia coli one of three smooth muscle bands that make up the longitudinal muscle layer of the muscularis in all of the large intestine except the terminal end
tongue accessory digestive organ of the mouth, the bulk of which is composed of skeletal muscle
transverse colon part of the colon between the ascending colon and the descending colon
upper esophageal sphincter skeletal muscle sphincter that regulates food movement from the pharynx to the esophagus
Valsalva’s maneuver voluntary contraction of the diaphragm and abdominal wall muscles and closing of the glottis, which increases intra-abdominal pressure and facilitates defecation
villus projection of the mucosa of the small intestine
Contributors and Attributions
- Glossary: The Digestive System. Authored by : Kelly, Jessica. Provided by : Herkimer College. Located at : NA. Project : AtD Course. License : CC BY: Attribution


6.1: The Impacts of Environmental Conditions
Our industrialized society dumps huge amounts of pollutants and toxic wastes into the earth’s biosphere without fully considering the consequences. Such actions seriously degrade the health of the earth’s ecosystems, and this degradation ultimately affects the health and well-being of human populations.
For most of human history, biological agents were the most significant factor in health. These included pathogenic (disease causing) organisms such as bacteria, viruses, protozoa, and internal parasites. In modern times, cardiovascular diseases, cancer, and accidents are the leading killers in most parts of the world. However, infectious diseases still cause about 22 million deaths a year, mostly in undeveloped countries. These diseases include: tuberculosis, malaria, pneumonia, influenza, whooping cough, dysentery and Acquired Immune Deficiency Syndrome (AIDS). Most of those affected are children. Malnutrition, unclean water, poor sanitary conditions and lack of proper medical care all play roles in these deaths. Compounding the problems of infectious diseases are factors such as drug-resistant pathogens, insecticide resistant carriers, and overpopulation. Overuse of antibiotics have allowed pathogens to develop a resistance to drugs. For example, tuberculosis (TB) was nearly eliminated in most parts of the world, but drug-resistant strains have now reversed that trend. Another example is malaria. The insecticide DDT (a chemical called dichlorodiphenyltrichloroethane) was widely used to control malaria-carrying mosquito populations in tropical regions. However, after many years the mosquitoes developed a natural resistance to DDT and again spread the disease widely. Anti-malarial medicines were also over-prescribed, which allowed the malaria pathogen to become drug-resistant.
Chemical agents also have significant effects on human health. Toxic heavy metals, dioxins, pesticides, and endocrine disrupters are examples of these chemical agents. Heavy metals (e.g., mercury, lead, & cadmium) are typically produced as by-products of mining and manufacturing processes. All of them biomagnify (become more concentrated in species with increasing food chain level). For example, mercury from polluted water can accumulate in swordfish to levels toxic to humans. When toxic heavy metals get into the body, they accumulate in tissues and may eventually cause sickness or death. Studies show that people with above-average lead levels in their bones have an increased risk of developing attention deficit disorder and aggressive behavior. Lead can also damage brain cells and affect muscular coordination.
ENVIRONMENTAL PERSISTENCE OF DDT
The pesticide DDT was widely used for decades. It was seen as an ideal pesticide because it is inexpensive and breaks down slowly in the environment. Unfortunately, the latter characteristic allows this chemical agent to biomagnify through the food chain. Populations of bird species at the top of the food chain, e.g., eagles and pelicans, are greatly affected by DDT in the environment. When these birds have sufficient levels of DDT, the shells of their eggs are so thin that they break, making reproduction impossible. After DDT was banned in the United States in 1972, affected bird populations made noticeable recoveries, including the iconic bald eagle.
Contributors and Attributions
- Essentials of Environmental Science by Kamala Doršner is licensed under CC BY 4.0