 ideal for them.
As there are no rear half central laterals installed yet, it's why there's little "curve" to it at the moment. Those laterals will push it out further in the middle later..
So tomorrow start on the front halves..
Went in this afternoon and put the tail post on, adding some bracing inside the rear half, and stuck the front halves and firewall on...
So this morning I had the side panels lasered and folded. Bit hard to see but all the pilot holes are there so you just pop the drill through without measuring or mucking around.
The panels fit snugly so you just simply line them up, push them on, then drill away. All these panels are sitting their on their own.
All the measuring, scribing and pin punching work has been completely eliminated, and of course the laser is deadly accurate and makes left and right sides alignment near perfect.
Nothing overly special today, just fitting up the turtle deck and tail to the new sides and some internal bracing in the tail...
Over the weekend I redesigned the wing box for some good reasons, as well as ever, simplicity, speed and cost.
Yesterday and today busy building it.
The theory is simple, tapered wing ends into a tapered hole, self locating and aligning. I just got enough of one partial wing finished tonight and it works a treat, fits so **** good that I actually am concerned about taper stiction, I had to wiggle it quite a bit to get it back out!
.. and also putting the front half back onto the rear half, note the carry through tubes that someone was concerned about. Also note there is a substantial outer joining plate not shown yet.
.. and time to whip up a couple of flappy arms for it.
So this is what the wing is looking like (in progress), note I use a folded C channel shear web, besides a little added strength, it's main function is as a jigging entity that keeps the assembly dead straight...
Annoyingly the rear turtle deck front skin was a little too short (25mm/1") and I had to make a new one. So I decided to make it just one piece and rivet to the spine as you can see here.
Good thing is the entire rear turtle decks and tail can be assembled on a flat bench or floor in one piece as you see here then simply added to the frame as the one piece, of course I've set it up so that's all also self jigging. It's so darn easy.
Just another step closer, slowly chipping away at it...
For those curious, I have been on holidays for the last 3 weeks in 90 to 110 degree heat wave in Australia.
Start back tomorrow on this.
So I threw the sides on today, gearing up for a fuselage load test sometime soon...
So getting the rear horizontal stabiliser and elevator underway..
The HS is coming together bit by bit...# Vectors

A vector describes a movement from one point to another. It is a mathematical quantity having both Magnitude & Direction. The length of the segment of the directed line is called the magnitude of the vector and the angle at which the vector is inclined shows the direction of the vector.

The beginning point of a vector is called as “Tail” and the end side (having arrow) is called as “Head.”

### Examples & Representation-

Velocity, Acceleration, Force, Increase/Decrease in Temperature etc.

A vector between two points A and B is given as $\overrightarrow{AB}$, or vector a.

Breaking a vector into its x and y components is the most common way for solving vectors.

A vector “a” is inclined with horizontal having an angle equal to $\theta$.

This given vector “a” can be broken down into two components i.e. ax and ay.

The component ax is called as “Horizontal component” whose value is $a \cos \theta$.

The component ay is called as “Vertical component” whose value is $a \sin \theta$.

 Example- Given vector V, having magnitude of 10 units & inclined at $60^{\circ}$. Break down the given vector into its two component. Solution- $\overrightarrow{v}$, having magnitude(V) = 10 units and $\theta = 60^{\circ}$ Horizontal component (Vx) = $V \cos \theta$ Vx = $10\; \cos 60^{\circ}$ Vx = $10 \times 0.5$ Vx = 5 units Now, Vertical component(Vy) = $V \sin \theta$ Vy = $10\; \sin 60^{\circ}$ Vy = $10 \times \frac{\sqrt{3}}{2}$ Vy = $10 \sqrt{3}$ units

## Magnitude of a Vector-

The magnitude of a vector is shown by vertical lines on both the sides of the given vector.

$\left | a \right |$

Mathematically, the magnitude of a vector is calculated by the help of “Pythagoras Theorem,” i.e.

$\left | a \right | = \sqrt{x^{2}+y^{2}}$,

 Example- Find the magnitude of vector a (3,4). Solution- Given $\overrightarrow{a}$= (3,4) $\left | a \right |= \sqrt{x^{2}+y^{2}}$ $\left | a \right |= \sqrt{3^{2}+4^{2}}$ $\Rightarrow \left | a \right |= \sqrt{9 + 16} = \sqrt{25}$ Therefore, $\left | a \right |= 5$

## Operation on Vector–

Vector operation such as Addition, Subtraction, Multiplication etc. can be done easily.

The two vectors a and b can be added giving the sum to be a + b. This requires joining them head to tail.

We can translate the vector b till its tail meets the head of a. The line segment that is directed from the tail of vector a to the head of vector b is the vector “a + b”.

• Commutative Law- the order of addition does not matter, i.e, a + b = b + a
• Associative law- the sum of three vectors has nothing to do with which pair of the vectors is added at the beginning.

i.e. (a + b) + c = a + (b + c)

### 2. Subtraction of Vectors-

Before going to the operation it is necessary to know about reverse vector(-a).

A reverse vector (-a) which is opposite of a has similar magnitude as a but pointed in opposite direction.

First, we find the reverse vector.

Such as if we wanna find vector b – a

Then, b – a = b + (-a)

### 3. Multiplication of Vectors-

• #### Scalar Multiplication-

Multiplication of a vector by a scalar quantity is called “Scaling.”

In this type of multiplication, only the magnitude of a vector is changed not the direction.

• #### Vector Multiplication-

It is of two types “Cross product” and “Dot product.”

### Cross Product-

The cross product of two vectors results in a vector quantity. It is represented by a cross sign between two vectors.

 $a \times b$

Mathematical value of a cross product-

 $a \times b = \left | a \right | \left | b \right |\sin \theta \;\hat{n}$

where, $\left | a \right |$ is the magnitude of vector a.

$\left | a \right |$ is the magnitude of vector b.

$\theta$ is the angle between two vectors a & b.

and $\hat{n}$ is a unit vector showing the direction of the multiplication of two vectors.

### Dot product-

The dot product of two vectors always result in scalar quantity, i.e. it has only magnitude and no direction. It is represented by a dot in between two vectors.

 $a.b$

Mathematical value-

 $a. b = \left | a \right | \left | b \right |\cos \theta$
 Example- Find the scalar and vector multiplication of two vectors a and b given by $3\hat{i}-1\hat{j}+2\hat{k}$ and $1\hat{i}+-2\hat{j}+3\hat{k}$ respectively. Solution- Given vector a (3,-1,2) and vector b (1,-2,3) Vector product ( or Cross product) = $\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & -1 & 2 \\ 1 & -2 & 3 \end{vmatrix}$ $\vec{a} \times \vec{b} = \begin{vmatrix} -1 & 2 \\ -2 & 3 \end{vmatrix} \hat{i} – \begin{vmatrix} 3 & 2\\ 1 & 3 \end{vmatrix}\hat{j} + \begin{vmatrix} 3 & -1\\ 1 & -2 \end{vmatrix}\hat{k}$ $\vec{a} \times \vec{b} = 1 \hat{i} – (7)\hat{j} + (-5) \hat{