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Tutorial on a Fuel time/timer gauge

133 MrCacahuate  2.7 years ago

Hi, first a disclaimer, Im not 100% sure that what I did is correct, so if you have suggestions/corrections are very welcome.

¿What is this about?

A simple guide on how to code a fuel timer gauge, (how much time in hours, minutes, seconds, is left before you run out of fuel at a given fuel burn rate).

Code for the input:

Input: Fuel/(-rate(Fuel/X)*Y)

Where:
Fuel: is the amount of fuel remaining in the aircraft, if your tank is full it returns 1, if it is half empty then 0.5 and so on...
rate(A): returns the rate of change of A, since fuel decreases a minus is added.
X: fuel consumption of your reference engine, (if you have different engines or more than 1 engine, dont worry, just select which engine you want as a reference)
Y: is the amount of fuel in liters

Code for our range/multiplier:

For the gauge Im using the altimeter, and saying that each mark represents seconds10, minutes10 and hours*10, meaning that if I have the minute pointer between 5 and 4 I have 45 minutes remaining of fuel at the current fuel consumption rate.

The tricky part of this gauge is to correctly set the range, I'll try to explain my reasoning as best as I can.

Imagine that you have 75 liters or approx 20gal, if your fuel consumption rate is 4 liters per second, then you'll be out of fuel in 75/4 seconds (18.75s).

This means that I want the sec. hand to point exactly at the 1.875 mark on my gauge.

The maximum return value of my Input code will be "1", therefore, my maximum range should be "At how many degrees" my sec. hand points at 1.875, doing a simple rule of three we get:

(1.875*360)/10° = 67.5

If you want to setup the minutes and hours gauge simply divide the value of seconds by 60 to get minutes or by 3600 to get hours.

In resume, you need to do the following for any gauge:

Seconds:

multiplier in degrees = ((Amount of fuel in liters/fuel consumption)*(maximum gauge range))/(max face value))/(Your Scale)

Minutes:

multiplier in degrees = (((Amount of fuel in liters/fuel consumption)/60)*(maximum gauge range))/(max face value))/(Your Scale)

Hours:

multiplier in degrees = (((Amount of fuel in liters/fuel consumption)/3600)*(maximum gauge range))/(max face value))/(Your Scale)

Worked example:

You've designed an aircraft that has the following characteristics:
3 X Blasto BF150
2 X Blasto BF300
Fuel = 8671 gal
You want to design a fuel timer gauge using the altimeter gauge containing only readings for minutes and hours of flight remaining

1.- Select a reference engine
Choose either the Blasto BF150 or the Blasto BF300 and find their fuel consumption rates, in this case are 4 l/s and 4.5 l/s respectively for each one; in this case I've chosen the blasto BF150 as my reference but it should be the same for any engine.

2.-Calculate your fuel.
If you dont have the exact value of your fuel in liters, you can find and approximate by doing: (Amount of fuel in gal/20)*75, this is a simple rule of three derived from the fact that 75 liters in simple planes in approx 20 gal. Doing this operation yields a value of 32516.25 liters, (the actual value is 32823.75, an error of less than 1%)

3.-Set the input for the minutes hand, this is:
Fuel/(-rate(Fuel/X)*Y) = Fuel/-rate(Fuel/4)*32823.75

3.5.- Set the range/multiplier of minutes hand as:
(((Amount of fuel in liters/fuel consumption)/60)*(maximum gauge range))/(max face value))/(Your Scale) = (((32823.75/4)/60)*360)/(10))/10

4.- Set the input for the hours hand:
Fuel/(-rate(Fuel/X)*Y) = Fuel/-rate(Fuel/4)*32823.75

4.5.- Set the range/multiplier of the hour hand:
(((Amount of fuel in liters/fuel consumption)/3600)*360)/(max face value))/(Your Scale) = (((32823.75/4)/3600)*360)/(10))/10

5.- Dicussion of results:
If only a single Blasto BF150 engine where to be operational with a throttle of 100%, our aircraft will be left out of fuel after 136.76 minutes, (our multiplier should be 492.35°), or 2.26 hours, (our multiplier should be 8.206°).

Now, imagine that all engines are running at 100% throttle, this means our fuel consumption is: (4 X 3) + (4.5 X 2) = 21 l/s, and our endurance then becomes: 0.4341 hours or 26 minutes.

Disclaimers

1.-For low values of the fuel consumption the gauge works descently, however, as the liters increase, the gauge starts to act wonky
2.- Slowmotion readings are innacurate

Suggestions and corrections are welcomed n.n