# How to Express This as a Simple Equation

**URL:** <https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443>\
**Category:** Factual Questions\
**Created:** [March 11, 2014, 3:32pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443 "2014-03-11T15:32:15Z")\
**Posts on this page:** 9\
**Page:** 1

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**Author:** ![HeyHomie](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/heyhomie/32/207_2.png) [@HeyHomie](https://boards.straightdope.com/u/HeyHomie)\
**Post date:** [March 11, 2014, 3:32pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/1 "2014-03-11T15:32:15Z")

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What is the equation for when an increase in one value necessitates an increase in another value? If I’m not being clear (and I reckon I’m not), here’s a real life example:

An aircraft weighs, say, 16000 kg when full of passengers & crew, luggage, etc. To move a 16000 kg craft from Departure Point to Destination Point requires, say, 5000 liters of fuel (\*pulling that number out of my ass; I have no idea how much fuel a jet consumes). 5000 liters of fuel adds another 4000 kg to the weight of the craft, so now it weighs 20000 kg. So you need more fuel to account for the weight of the fuel itself; you will then need more fuel to account for the weight of the additional fuel; and so on.

How can this be expressed in one simple equation?

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**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\
**Post date:** [March 11, 2014, 4:09pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/2 "2014-03-11T16:09:09Z")

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I’d start with:

Fuel required = 5000 + (Fuel required / 4)

The 5000 being required to lift the rest of the plane, and assuming that 1 liter of fuel can always serve to lift 4 liters of fuel. You may need to make the term more complicated if it’s not a simple relationship like that.

Then you can solve for Fuel required:

0.75 \* Fuel required = 5000  
Fuel Required = 5000 / 0.75  
Fuel Required = 6666.67 approx

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**Author:** ![bob\_2](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bob_2/32/3341_2.png) [@bob\_2](https://boards.straightdope.com/u/bob_2)\
**Post date:** [March 11, 2014, 4:23pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/3 "2014-03-11T16:23:35Z")

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Pretty easy to construct a graph I would have thought.

It’s like one of those problems where you have to send trucks into the desert but they can’t carry enough fuel for the journey. You have to send truck one, x miles to make a fuel dump, and then return. ruck two goes further and makes a second dump, and so on. The challenge is to make the fewest number of trips and use the least amount of fuel for a given mileage and truck capacity.

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**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\
**Post date:** [March 11, 2014, 4:30pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/4 "2014-03-11T16:30:02Z")

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> [@HeyHomie](#):
>
> 5000 liters of fuel adds another 4000 kg to the weight of the craft, so now it weighs 20000 kg. So you need more fuel to account for the weight of the fuel itself; you will then need more fuel to account for the weight of the additional fuel; and so on.

This sounds similar to compound interest: you earn interest on the money in the account, but then you need to add on the interest earned on that interest, plus the interest earned on _that_ interest…

This would be an example of exponential growth, which would be modeled by an equation of the form y = Ce[sup]kx[/sup]; but other factors, such as the weight of the passengers and crew, may make the equation more complicated.

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**Author:** ![HeyHomie](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/heyhomie/32/207_2.png) [@HeyHomie](https://boards.straightdope.com/u/HeyHomie)\
**Post date:** [March 11, 2014, 4:48pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/5 "2014-03-11T16:48:44Z")

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> [@Thudlow\_Boink](#):
>
> This would be an example of exponential growth, which would be modeled by an equation of the form y = Ce[sup]kx[/sup];

That’s what I was looking for, thanks!

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**Author:** ![txtumbleweed](https://avatars.discourse-cdn.com/v4/letter/t/278dde/32.png) [@txtumbleweed](https://boards.straightdope.com/u/txtumbleweed)\
**Post date:** [March 11, 2014, 5:27pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/6 "2014-03-11T17:27:21Z")

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There are two equations I remember running across that might apply. I am a liberal arts major, not a math major, so hopefully someone with actual experience with these equations will be along to help translate them.

For aircraft, you can use the [Breguet Range Equation](http://virtualskies.arc.nasa.gov/aeronautics/7.html).

> [@](#):
>
> To find the range is much more complex than simply dividing the total fuel by the average hourly fuel consumption and then multiplying by the velocity. That’s why we need the Breguet Range equation to predict how far an airplane will fly accounting for flight performance and the change in weight of the airplane as it burns or uses its fuel.  
> The Breguet Range equation is given below.  
> Range = (V/C) \* (L/D) \* ln(Wi/Wf)  
> V= Velocity in knots  
> C= Specific fuel consumption in pounds fuel per pounds thrust per hour  
> L= Lift in pounds  
> D= Drag in pounds  
> ln= Natural logarithm, or loge  
> Wi= Weight of aircraft at start of cruise in pounds  
> Wf= Weight of aircraft at end of cruise in pounds

For rockets, the [Tsiolkovsky Rocket Equation](http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) describes the motion of a vehicle that expels part of its mass to accelerate.

[Relevant XKCD](https://what-if.xkcd.com/7/)

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**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\
**Post date:** [March 11, 2014, 5:33pm UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/7 "2014-03-11T17:33:05Z")

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> [@Thudlow\_Boink](#):
>
> This would be an example of exponential growth

I’m not a math wiz, but I can’t see how this can possibly be _exponential_ growth. Adding 400 liters of fuel to the plane makes its fuel requirement increase exponentially? That just doesn’t sound right.

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**Author:** ![Zero-syde](https://avatars.discourse-cdn.com/v4/letter/z/a3d4f5/32.png) [@Zero-syde](https://boards.straightdope.com/u/Zero-syde)\
**Post date:** [March 12, 2014, 8:38am UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/8 "2014-03-12T08:38:44Z")

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> [@KneadToKnow](#):
>
> I’m not a math wiz, but I can’t see how this can possibly be _exponential_ growth. Adding 400 liters of fuel to the plane makes its fuel requirement increase exponentially? That just doesn’t sound right.

Considering it looked like a geometric series to me (without doing any computations, just a glance) I suspect he means exponential decay. The two (decay and growth) are basically the same thing, so he’s not wrong, but k \< 1

Something like adding x lbs weight requires y lbs fuel, where y is x/2 or something, means you’re adding x + x/2 + x/4 + x/8 + … which is exponential (look at the denominator). But I would never try to solve it that way. It’s a simple system of equations, I think.

I suspect this will match an earlier post but lets just do this anyway:

total\_weight = fuel\_weight + initial\_weight  
fuel\_weight = total\_weight_gallons\_per\_pound_weight\_per\_gallon

I tried to avoid ambiguity but gallons\_per\_pound (GPP) is the amount of gallons required to lift 1 pound, and weight\_per\_gallon (WPG) is the actual weight of a gallon of fuel. The logic here is that we _know_ the total weight is the initial+the fuel, and we also know that the weight of the fuel is capable of lifting the total weight, so all I did was back convert. Now, solving the system…

total\_weight = total\_weight_GPP_WPG + initial\_weight  
total\_weight - total\_weight_GPP_WPG = initial\_weight  
total\_weight\*(1-GPP\*WPG) = initial weight

total\_weight = initial\_weight/(1-GPP\*WPG)

In the problem given, WPG =0.8, GPP = 0.3125 and initial\_weight = 16000kg thus total\_weight = 21333.3 kg, fuel\_weight = 5333.3 kg and the number of liters is 6666.7L.

The answer, in one equation then, would be **total\_weight = initial\_weight/(1-liters\_to\_lift\_a\_kilogram\*kilograms\_per\_liter)**

—Edit: I should have just simplified this to total\_weight = initial\_weight/0.75, as the long names are really constants provided in the initial problem!—

Note: I was halfway through the post when I realized that you actually _weren’t_ using crappy imperial units (great irony, as I’m American and they make no sense to me). But I didn’t change it because the 3 letter abbreviation is easier in imperial units.

Note: The 1-x term scared me (what if x\>1?) but it makes sense. If it takes more than a kilogram of fuel to lift a kilogram, you’re going to have negative values!

Note: This is only a solution to the problem as given, and doesn’t take into account the compounding effects of, ultimately, F = MA (where burned fuel is generating a relatively constant F, and A depends on the M of the plane). The formulas given by a previous poster is a MUCH more accurate model of reality.

Note: I suck at math, and probably screwed this up.

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**Author:** ![Zero-syde](https://avatars.discourse-cdn.com/v4/letter/z/a3d4f5/32.png) [@Zero-syde](https://boards.straightdope.com/u/Zero-syde)\
**Post date:** [March 12, 2014, 8:51am UTC](https://boards.straightdope.com/t/how-to-express-this-as-a-simple-equation/683443/9 "2014-03-12T08:51:34Z")

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Hmmm. I think I focused on the “problem” without answering the OP’s question. Here’s a better response:

It depends. The relationship between the values determines how the solution will be found. Often this will involve a system of equations, but that system may not produce results that are as easily separated as the example above.

Calculus _is_ the study of related rates of change. Linear Algebra delves quite deeply into variables that depend on variables (systems of equations), and so the combined toolkit of Calculus and Linear Algebra is probably what you’re going to need to analyze and solve most problems like this.

There is no single equation that can solve all situations, but there are approaches that are pretty darn good at finding specific equations that solve specific situations.