# Weight and Balance "constant" question

Thread Starter

Join Date: Jul 2009

Location: Earth

Posts: 102

**Weight and Balance "constant" question**

Hello there.

So, the way I learned Weight and Balance problems, was that for a given station weight, there was a given moment. And in the POH there are the graphs for making it easier to find it. And in other manuals there are the tables for, i.e. given fuel weight has a given moment.

However I can't work this out: I have some W&B spreadsheets for some light aircraft, and in the fuel field (for example, could also be baggage or pilot and passengers) there is a constant the author used which I think represents the formula for the loading weight / loading moment line graph.

Here is the graph in question:

In this case, a C152, the sheet multiplies the fuel weight (in pounds) with 0.04217.

If I just did the basic multiplication of weight times the arm of the fuel tank station, I wouldn't find the same answer, since it seems that with added weight the arm will gradually increase (or am I wrong?)

What I want to know is where did this "constant" of 0.04217 (in this case) come from?

I know that for different aircraft, and different stations (say fuel, pax, crew, etc.) it would be different constants, but I just can't find it in any manual I come across.

I may be missing something here, assuming something wrong.

Anyone could help me on this?

Thanks

So, the way I learned Weight and Balance problems, was that for a given station weight, there was a given moment. And in the POH there are the graphs for making it easier to find it. And in other manuals there are the tables for, i.e. given fuel weight has a given moment.

However I can't work this out: I have some W&B spreadsheets for some light aircraft, and in the fuel field (for example, could also be baggage or pilot and passengers) there is a constant the author used which I think represents the formula for the loading weight / loading moment line graph.

Here is the graph in question:

In this case, a C152, the sheet multiplies the fuel weight (in pounds) with 0.04217.

If I just did the basic multiplication of weight times the arm of the fuel tank station, I wouldn't find the same answer, since it seems that with added weight the arm will gradually increase (or am I wrong?)

What I want to know is where did this "constant" of 0.04217 (in this case) come from?

I know that for different aircraft, and different stations (say fuel, pax, crew, etc.) it would be different constants, but I just can't find it in any manual I come across.

I may be missing something here, assuming something wrong.

Anyone could help me on this?

Thanks

*Last edited by Pugachev Cobra; 13th May 2010 at 16:25.*

Join Date: Feb 2006

Location: Sweden

Posts: 42

If you ask me: For 100 pounds of fuel the moment would be close to 3800 pound-inches. This seems to hold throughout the table (straight line)

This means that 1 pound gives the moment 38 pound-inches = arm ~38 inches.

Could it not be that the arm of the fuel cg is 38 inches from the datum? Or was the question about something else?

This means that 1 pound gives the moment 38 pound-inches = arm ~38 inches.

Could it not be that the arm of the fuel cg is 38 inches from the datum? Or was the question about something else?

Thread Starter

Join Date: Jul 2009

Location: Earth

Posts: 102

Yes I can see we can derive that the fuel tank cg station is near station 38.

However, and this puzzles me a bit, I have 3 manuals here for the C152, C172 and C182, and I see no reference to the fuel tanks cg arm. Cessna states that you don't need to use the cg arm on the "loading problem", you just use the graph, which of course works fine.

But I really think somewhere we should have the arm for the fuel tanks. And where exactly did the author of those W&B spreadsheets find that constant value to multiply the weight?

Different constants for different positions. I'm just curious about that.

I realized this missing fuel tank arm issue the other day, when I tried to calculate the CG without using the graph. I had no information for the fuel tank arm. So I went to the spreadsheet and found the constant, which is even more mysterious to me at this point.

However, and this puzzles me a bit, I have 3 manuals here for the C152, C172 and C182, and I see no reference to the fuel tanks cg arm. Cessna states that you don't need to use the cg arm on the "loading problem", you just use the graph, which of course works fine.

But I really think somewhere we should have the arm for the fuel tanks. And where exactly did the author of those W&B spreadsheets find that constant value to multiply the weight?

Different constants for different positions. I'm just curious about that.

I realized this missing fuel tank arm issue the other day, when I tried to calculate the CG without using the graph. I had no information for the fuel tank arm. So I went to the spreadsheet and found the constant, which is even more mysterious to me at this point.

Join Date: Nov 1999

Location: UK

Posts: 61

Well, unless I'm missing something it seems fairly straight forward.

The lines on the graph all appear to be straight so the arms must therefore all be constant. The fuel appears to have an arm of some 38.75", baggage area 1 some 64" and baggage area 2 some 85.7". And if I read the graph correctly, the occupants and the fuel have the same arm.

As weight and balance seems to be a bit of a black art to many, the manufacturer has probably chosen to present the data in this fashion to simplify the issue as much as possible, thus increasing the chances that pilots will do w&b calculations at all whilst at the same time reducing the chances of gross errors.

Even though the arm data might not appear in the AFM, the author of the graph will have obtained it from the manufacturer.

For simple aircraft such as this the fuel arm will generally be constant (a rectangular fuel tank fitted in a straight wing). However on swept wing aircraft which may also have belly tanks and then some sort of automatic scheduling system for the order in which tanks are used, a graph of the fuel arm can be a very convoluted thing.

Hope it helps.

MT

The lines on the graph all appear to be straight so the arms must therefore all be constant. The fuel appears to have an arm of some 38.75", baggage area 1 some 64" and baggage area 2 some 85.7". And if I read the graph correctly, the occupants and the fuel have the same arm.

As weight and balance seems to be a bit of a black art to many, the manufacturer has probably chosen to present the data in this fashion to simplify the issue as much as possible, thus increasing the chances that pilots will do w&b calculations at all whilst at the same time reducing the chances of gross errors.

Even though the arm data might not appear in the AFM, the author of the graph will have obtained it from the manufacturer.

For simple aircraft such as this the fuel arm will generally be constant (a rectangular fuel tank fitted in a straight wing). However on swept wing aircraft which may also have belly tanks and then some sort of automatic scheduling system for the order in which tanks are used, a graph of the fuel arm can be a very convoluted thing.

Hope it helps.

MT

Thread Starter

Join Date: Jul 2009

Location: Earth

Posts: 102

It does help Mach Tuck.

However, seems to me I wasn't clear enough.

The graph is from the AFM/POH. The author of the graph is Cessna.

What I don't get is the constant 0.04217 for the fuel tanks, that someone used in an Excel Spreadsheet, to mimic the graph.

You just enter the gallons, the spreadsheet multiplies by 6 (6 lbs / gal) and then multiplies again by 0.04217. And it gets the precise moment.

Now, where did the number 0.04217 come from? (Not just this number, but any number...)

I can't really buy the idea that 2 different persons just derived mathematically this value by applying some unknown formula from the AFM's graph.

If it helps, the baggage area 1 and 2 respectively have the 'constants' 0.06346 and 0.08666.

The AFM does give us the stations for the baggage areas, but in this spreadsheet they aren't used at all.

But from your answers, I can see that this "constant" number isn't really a part of official documentation, AFM/POH etc, right?

However, seems to me I wasn't clear enough.

The graph is from the AFM/POH. The author of the graph is Cessna.

What I don't get is the constant 0.04217 for the fuel tanks, that someone used in an Excel Spreadsheet, to mimic the graph.

You just enter the gallons, the spreadsheet multiplies by 6 (6 lbs / gal) and then multiplies again by 0.04217. And it gets the precise moment.

Now, where did the number 0.04217 come from? (Not just this number, but any number...)

I can't really buy the idea that 2 different persons just derived mathematically this value by applying some unknown formula from the AFM's graph.

If it helps, the baggage area 1 and 2 respectively have the 'constants' 0.06346 and 0.08666.

The AFM does give us the stations for the baggage areas, but in this spreadsheet they aren't used at all.

But from your answers, I can see that this "constant" number isn't really a part of official documentation, AFM/POH etc, right?

Moderator

Join Date: Apr 2001

Location: various places .....

Posts: 6,517

First off you can download the C150 TCDS from the FAA site which gives the "official" loading arms. Never mind that the loading sheet graph doesn't quite match up with the TCDS data ...

If you're running with the normal ICAO/GAMA proforma (as with the chart cited) you are calculating the moment sum and then using the CG graph to figure the final CG. Just saves you the hassle of doing the sums longhand and there is no need,

Never mind sweptwing jets, any non-prismatic tank can be expected to have a fuel loading arm which varies with fuel volume - starting with small aircraft such as single Beech etc. The secret is to assess at what stage the aircraft is not reasonable for a linearised fuel loading arm.

Now, as to the "constants", there is no great mystery here.

By way of working up to the answer, consider converting a quantity to another quantity by a process of multiplication and division by several items eg -

2 x (6 x 8 / 2) = 48

can be done more economically by doing part of the intervening calculation and using simplified multipliers -

6 x 8 / 2 = 6 x 4 = 24

so we could have done the first sum as

2 x 6 x 4 = 48

and this is what the designer of the spreadsheet has done. Now why he/she did it that way and not go the whole bit and use 2 x 24 remains a mystery ... Perhaps he is an olde pharte like me and others who were brought up on old mainframes and limited significant places in floating point calculations .. which made us use simplified calculations on computers to minimise floating point calculation round off errors in consecutive calculations ? This was a particular problem if one had to do divisions with similar value quantities. Now, if that nonsense doesn't mean much to you, don't worry, it isn't anywhere near as important in these days of fancy PCs.

So, we can play with the numbers and come up with -

(a) for fuel

moment = pounds x arm / 1000

= USG x 6 x 42 /1000

= USG x 6 x 0.042

[More generally, one would refer to IU rather than moment in this case, but the subtlety is not important for the present discussion]

The spreadsheet designer either has used an arm of 42.17 (for whatever obscure reasons) or, more likely, he/she figured that value by doing a series of unit conversions. Either way, 0.04217 is just a minor simplification of the calculation - no more, no less. If you were working from the graph, with an arm of around 38, you would get 0.038, which is pretty close to 0.04217 at the end of the day.

(b) for the loading bays (and using Mach Tuck's read off values to save me the eye strain of doing it myself ..)

0.06346 = 63.46/1000 which is pretty close to the value derived from the graph of

0.064 = 64/1000

and

0.08666 = 86.66/1000 which is pretty close to the graph value of

0.0857 = 85.7/1000

It is not immediately obvious where the small discrepancies in the arms derive. However, if you like to post the spreadsheet, or send me a copy, it will become very obvious, very quickly, what the designer was up to.

which is a great shame as it is dead simple .. provided one follows the basics and maintains strict numerical housekeeping ... the black art comes into play when designing trimsheets and, even then, it's a simple exercise once you get the hang of how to go about it.

The manufacturer simply is following the recommended practices in the GAMA format POH. In my view, the GAMA format for weight and balance is dreadful but that's just my opinion. The GAMA format also has been adopted as an ICAO recommended light aircraft POH format so I guess I am just a lone voice in the wilderness on this point.

The TCDS is the first port of call and, at the end of the day, if we can't find out by other means, we get the old tape measure out and measure arms directly in the aircraft.

Now, I don't have a copy of the POH but let me guess, the quoted arms are pretty close to 63.46 and 86.66 ?

*I see no reference to the fuel tanks cg arm*If you're running with the normal ICAO/GAMA proforma (as with the chart cited) you are calculating the moment sum and then using the CG graph to figure the final CG. Just saves you the hassle of doing the sums longhand and there is no need,

*per se*, to know the actual loading arm values.*a graph of the fuel arm can be a very convoluted thing.*Never mind sweptwing jets, any non-prismatic tank can be expected to have a fuel loading arm which varies with fuel volume - starting with small aircraft such as single Beech etc. The secret is to assess at what stage the aircraft is not reasonable for a linearised fuel loading arm.

Now, as to the "constants", there is no great mystery here.

By way of working up to the answer, consider converting a quantity to another quantity by a process of multiplication and division by several items eg -

2 x (6 x 8 / 2) = 48

can be done more economically by doing part of the intervening calculation and using simplified multipliers -

6 x 8 / 2 = 6 x 4 = 24

so we could have done the first sum as

2 x 6 x 4 = 48

and this is what the designer of the spreadsheet has done. Now why he/she did it that way and not go the whole bit and use 2 x 24 remains a mystery ... Perhaps he is an olde pharte like me and others who were brought up on old mainframes and limited significant places in floating point calculations .. which made us use simplified calculations on computers to minimise floating point calculation round off errors in consecutive calculations ? This was a particular problem if one had to do divisions with similar value quantities. Now, if that nonsense doesn't mean much to you, don't worry, it isn't anywhere near as important in these days of fancy PCs.

So, we can play with the numbers and come up with -

(a) for fuel

moment = pounds x arm / 1000

= USG x 6 x 42 /1000

= USG x 6 x 0.042

[More generally, one would refer to IU rather than moment in this case, but the subtlety is not important for the present discussion]

The spreadsheet designer either has used an arm of 42.17 (for whatever obscure reasons) or, more likely, he/she figured that value by doing a series of unit conversions. Either way, 0.04217 is just a minor simplification of the calculation - no more, no less. If you were working from the graph, with an arm of around 38, you would get 0.038, which is pretty close to 0.04217 at the end of the day.

(b) for the loading bays (and using Mach Tuck's read off values to save me the eye strain of doing it myself ..)

0.06346 = 63.46/1000 which is pretty close to the value derived from the graph of

0.064 = 64/1000

and

0.08666 = 86.66/1000 which is pretty close to the graph value of

0.0857 = 85.7/1000

It is not immediately obvious where the small discrepancies in the arms derive. However, if you like to post the spreadsheet, or send me a copy, it will become very obvious, very quickly, what the designer was up to.

*As weight and balance seems to be a bit of a black art to many*which is a great shame as it is dead simple .. provided one follows the basics and maintains strict numerical housekeeping ... the black art comes into play when designing trimsheets and, even then, it's a simple exercise once you get the hang of how to go about it.

*the manufacturer has probably chosen to present the data in this fashion to simplify the issue as much as possible,*The manufacturer simply is following the recommended practices in the GAMA format POH. In my view, the GAMA format for weight and balance is dreadful but that's just my opinion. The GAMA format also has been adopted as an ICAO recommended light aircraft POH format so I guess I am just a lone voice in the wilderness on this point.

*the author of the graph will have obtained it from the manufacturer.*The TCDS is the first port of call and, at the end of the day, if we can't find out by other means, we get the old tape measure out and measure arms directly in the aircraft.

*The AFM does give us the stations for the baggage areas, but in this spreadsheet they aren't used at all.*Now, I don't have a copy of the POH but let me guess, the quoted arms are pretty close to 63.46 and 86.66 ?

Join Date: Nov 1999

Location: UK

Posts: 61

I had just formulated my reply when I noticed johntullamarine's excellent opus had beaten me to it. I'll post it anyway:

Using the conventional mathematical calculation method for working out C of G one can finish up with very large-number moment values. These numbers are often divided by 1,000 to make them more manageable (as at the bottom of your graph).

The author of the spreadsheet has chosen to follow this convention and has created his 'constants' by simply dividing the arm by 1,000 rather than the resultant moment (the answer, of course, is the same).

The fuel 'constant' is the fuel arm/1,000 factored by 6 because you are entering a value in gallons, not pounds. The spreadsheet does not, as you might expect, turn your gallons into pounds and then multiply by the arm, instead it's simply frigged the arm so that gallon values can be computed directly.

MT

Using the conventional mathematical calculation method for working out C of G one can finish up with very large-number moment values. These numbers are often divided by 1,000 to make them more manageable (as at the bottom of your graph).

The author of the spreadsheet has chosen to follow this convention and has created his 'constants' by simply dividing the arm by 1,000 rather than the resultant moment (the answer, of course, is the same).

The fuel 'constant' is the fuel arm/1,000 factored by 6 because you are entering a value in gallons, not pounds. The spreadsheet does not, as you might expect, turn your gallons into pounds and then multiply by the arm, instead it's simply frigged the arm so that gallon values can be computed directly.

MT

Thread Starter

Join Date: Jul 2009

Location: Earth

Posts: 102

Now, I don't have a copy of the POH but let me guess, the quoted arms are pretty close to 63.46 and 86.66 ?

= USG x 6 x 42 /1000

Never mind sweptwing jets, any non-prismatic tank can be expected to have a fuel loading arm which varies with fuel volume - starting with small aircraft such as single Beech etc. The secret is to assess at what stage the aircraft is not reasonable for a linearised fuel loading arm.

But as you put it, it doesn't apply to light aircraft such as C152's, 172's, 182's right? Since their fuel tanks are "standard" non-prismatic.

I actually thought of using the POH cabin arrangements section, which lists various stations from the firewall to the back of baggage area 2, to figure out the fuel tank station, since in this case the tanks are directly on the side of the pilot's seat. But in that case, the arm would be 39 (the seat stations range are from 33 to 41), and I just thought it wouldn't be right to do it that way, since it was not going to be accurate.

But John, you said we could just use a tape to measure it. So it can be done? I don't know how I would figure out the exact centroid for the fuel tanks, and couldn't know if the fuel tank cg would shift with different volumes in it.

And Mach Tuck, yes I see it now. It seemed to me that too many decimal digits wouldn't be derived by someone, it had to be written in somewhere.

And thank you for the FAA TCDS site, I did not know it. Will dig deep into it.

Moderator

Join Date: Apr 2001

Location: various places .....

Posts: 6,517

*Just to be clear, you derived the 42 from the constant I gave you right?*

Nope, lifted it straight out of the TCDS

*you said we could just use a tape to measure it.*

That works fine for other than fuel tanks as you can see the loading position and measure its shape .. from which you can calculate a centroid (or take a stab at estimating it if you are in a hurry).

For nice boxy fuel tanks, likewise, not a problem.

For fancier fuel tanks, either

(a) get the drawing or do the tank measurements and then do the sums - generally not practical

(b) get a fuel tank and, with increasing water quantities (not as smelly and exciting as fuel) run some weighings and CG calculations to develop a table of CG against fuel volume

(c) do (b) with the aircraft - generally the accuracy is not adequate

(d) look it up in the POH fuel loading table and, if the CGs are not cited explicitly, reverse engineer the tabulated data to figure them out. This is pretty standard stuff when developing trimsheets.