user57901: I'm trying to build a battery operated hand lantern. It has the Arduino, an MPU6050 accelerometer, and 4 APA106 (WS2812B-like) LEDs.
If everything's powered through the Arduino's 5V pins, that's 80mA for the Arduino, 240mA for the 4 LEDs, and let's round up to 5 mA for the MPU6050.
If I'm using 5x AA batteries for 7.5V (at full charge) that should be 2.5V * 325mA of power wasted or just under a watt that'll be heating up the regulator? I'm not sure if that's huge or tiny or in between or what.
What's the most sane way to power this device for an absolute newbie?

user359: Use en external voltage regulator (go switch-mode if you care about battery life) that is adequately cooled.

user13327: 1W is near the limit of what a small LDO can handle. It will get considerably hot.
I would use 4 batteries instead of 5, and a 3.3V Arduino clone instead of the UNO, and power the LEDs directly from the batteries. You could also use your original UNO, feeding it with 6V on a 5V pin through a simple diode, which would drop it to something like 5.4..5.3V.

user2967: I suggest using a DC-DC buck/boost voltage regulator. Those can take a wide range of input voltages and put out an adjustable output voltage 
This one on Amazon, for example, can take an input of 3-40V and provide an output of 1.5-35V(Adjustable). It is a switching power supply, so it should not produce much waste heat. That one is rated at 2A, or 3A if you add a heat-sink.
You could feed the regulated 5V into the USB connector on your Arduino, and to your other components as well. (Just take a USB cable, cut it, and tie the red and black wires to ground and +5V, respectively. 
You might need a filtering capacitor to protect the Arduino from voltage variations as the load on the power supply changes. 

57901: <p>I'm trying to build a battery operated hand lantern. It has the Arduino, an MPU6050 accelerometer, and 4 APA106 (WS2812B-like) LEDs.</p>

<p>If everything's powered through the Arduino's 5V pins, that's 80mA for the <a href="https://forum.arduino.cc/index.php?topic=135872.msg1021818#msg1021818" rel="nofollow noreferrer">Arduino</a>, 240mA for the <a href="https://docs.audectra.com/guides/hardware/led-power-consumption/" rel="nofollow noreferrer">4 LEDs</a>, and let's round up to 5 mA for the <a href="https://store.invensense.com/datasheets/invensense/MPU-6050_DataSheet_V3%204.pdf" rel="nofollow noreferrer">MPU6050</a>.</p>

<p>If I'm using 5x AA batteries for 7.5V (at full charge) that should be 2.5V * 325mA of power wasted or just under a watt that'll be heating up the regulator? I'm not sure if that's huge or tiny or in between or what.<br>
What's the most sane way to power this device for an absolute newbie?</p>
 359: <p>Use en external voltage regulator (go switch-mode if you care about battery life) that is adequately cooled.</p>
 13327: <p>1W is near the limit of what a small LDO can handle. It will get considerably hot.</p>

<p>I would use 4 batteries instead of 5, and a 3.3V Arduino clone instead of the UNO, and power the LEDs directly from the batteries. You could also use your original UNO, feeding it with 6V on a 5V pin through a simple diode, which would drop it to something like 5.4..5.3V.</p>
 2967: <p>I suggest using a DC-DC buck/boost voltage regulator. Those can take a wide range of input voltages and put out an adjustable output voltage </p>

<p><a href="https://rads.stackoverflow.com/amzn/click/com/B00TMFKJ9Q" rel="nofollow noreferrer" rel="nofollow noreferrer">This one on Amazon</a>, for example, can take an input of 3-40V and provide an output of 1.5-35V(Adjustable). It is a switching power supply, so it should not produce much waste heat. That one is rated at 2A, or 3A if you add a heat-sink.</p>

<p>You could feed the regulated 5V into the USB connector on your Arduino, and to your other components as well. (Just take a USB cable, cut it, and tie the red and black wires to ground and +5V, respectively. </p>

<p>You might need a filtering capacitor to protect the Arduino from voltage variations as the load on the power supply changes. </p>
