View Full Version : Balance computation in %MAC

nobbyknownowt

8th Dec 2006, 20:48

:uhoh:

ARRRGGGGHHHHH!!!!

in the calculation a - (b*RA)/W where

a= distance of fulcrum (mainwheel jacking point)

b= distance difference between jacking points (nose and mainwheel)

RA = weight on nose wheel

W = total weight

we get the CG H arm

in the calculation c - (d*RA)/W

where

RA and W are the same as above

to get the CofG in %MAC

am sick of browsing the web

whats c & d???

cheers

Nobby

Mad (Flt) Scientist

8th Dec 2006, 21:37

if x is the reference point for the "leading edge" of the mac and "mac" is the length of the mac, then:

c=(a-x)/mac

d=b/mac

(which will give you cg as a fraction of mac; if you want it as a %age, then multiply then both by 100, i.e

c=100*(a-x)/mac

d=100*b/mac)

To prove this, consider that you already know:

h=a-(RA*b)/W

and if you know x and mac, then:

cg=(h-x)/mac

therefore

cg=( [a-(RA*b)/W] - x)/mac

= (a-x)/mac - [ (RA * b)/(W * mac) ]

= (a-x)/mac - [ (RA/W) * (b/mac) ]

so if cg = c - d RA/W, then c and d can be simply seen from the equation

john_tullamarine

8th Dec 2006, 22:16

The first calculation is simple enough to derive and is one of a number of ways of expressing the CG .. not one which I would use, but that is not relevant to the discussion, I guess.

The second appears to me to be so much arrant nonsense .. wherever did you come across it ? And, if you could be so kind to advise the answer .. did the person or book whence it came suggest any practical use for it ?

However, if one wanted to express the CG in %MAC values using this format, it is easy enough to derive the components by substituting the CG expression from the first post into the standard %MAC equation ... and then juggling the bits around a tad to come up with the required layout .. (careless mistakes on my part excepted ..)

LEMAC = distance from datum to leading edge of the MAC

MAC = length of the MAC

remainder as per the first post ...

%MAC = 100(CG - LEMAC)/MAC

= (100Wa - 100bRA - 100WLEMAC)/WMAC

= 100/MAC*(a - LEMAC) - 100b/MAC*RA/W

hence

c = 100/MAC*(a - LEMAC)

d = 100b/MAC

.... but who in their right senses would want to do such an apparently strange sort of thing ? My tiny little mind is reeling just from the suggestion ...

.. or have I missed something fundamental to weight control over the past 40-odd years ?

I see that MFS has responded at the same time ... might just as well leave both posts up, I guess.

Mad (Flt) Scientist

9th Dec 2006, 04:21

At least we both got the same answer. Worth preserving if only for the unusual event of two engineers giving the same answer ;)

john_tullamarine

9th Dec 2006, 07:21

.. mate .. when I saw that you had posted your response just ahead of mine .. I checked that that was the case .. be embarrassing if we couldn't get the same answer on a calculation as trivial as this one ....

nobbyknownowt

9th Dec 2006, 10:39

If I said a certain European aircraft consortiums STC manufacturer aircraft weighing reports. Would that be enough for you without me facing libel charges ?

although they call it "calcul du centrage avion en %mac"

When I came across it I could not fathom it as it just sits there, no explanation! and thought perhaps I'm missing something hence the large amount of web searching before finally posting here.

As I can use the arm and weight from the report I shall continue as before using my own methods.

cheers

Nobby

john_tullamarine

9th Dec 2006, 12:05

Absolutely no problem using the methodology if the good folk do it that way ... I just can't see any advantage in making a simple thing difficult. Given that none of us do it in our heads these days (more power, Excel), it really doesn't matter how the calculations are done, provided that the relationships are correct and we guard against numerical errors associated with the ways in which computers do their internal calculations.

nobbyknownowt

9th Dec 2006, 18:09

Thanks

Is there a decent resource out there that lists all these different methods so I could evaluate the most appropriate method?

Cheers

Nobby

john_tullamarine

9th Dec 2006, 19:36

Other than folk such as MFS and me (and many others), who play with this sort of stuff as a normal part of our day jobs, probably not much around other than specific in-house courses which various operators etc., have put together for their own use.

By all means, continue to raise questions and we can play with our pencils on paper to work out some answers for you .. part of the value of PPRuNe is the core of specialists who are seduced by fun in the sandpit and spend far more of their time here than they (probably) ought ...

nobbyknownowt

9th Dec 2006, 22:10

OK thanks it sure helps when I understand why something works.

theres an on wheels weighing formula that involves the sine of the pitch angle to get to the H-arm. Hows that work then??

cheers

Nobby.

john_tullamarine

10th Dec 2006, 00:48

Would you like to quote the formula and then we can comment ?

Some larger tailwheel aircraft have such approaches in the paperwork (DC3, eg). However, the problem lies in the presumption of a vertical position for the centre of mass.

In all cases, it is preferable to weigh the aircraft in the nominated level reference attitude to avoid this problem

nobbyknownowt

10th Dec 2006, 10:01

OK Thanks, By the way, dont you sleep??

Some background data. All from the same source

An aircraft on wheels weighing

Aircraft main wheels are at 12.871M

Nose wheel 4.089M

Main jack points at 12.763

Nose jack 4.426

Though I think thats irrellevant

LEMAC 11.425

MAC 2.285

(The fuselage datum lime is at 3M if thats any use)

a = Aircraft pitch attitude (+nose up attitude)

r = reaction on nose landing gear tyre

W = total wt

The H-arm M calc given is

12.908 - 0.686 sin a - (8.832 r / W)

The XG(%MAC) given is

64.9 - 30 sin a - (386.5 r/ W)

And if the aircraft has fuel then this is added into the equation as

Mfuel = Total fuel wt in Kg as indicated in #######

H-arm 12.908 - (0.686 + 0.00016 Mfuel) sin a - (8.832 r/W)

and

XG(%MAC) 64.9 - (30 + 0.007 Mfuel) sin a - (386.5 r / W)

Now obviously I can compute these equations and work from the results, but I'd love to understand what they are on about and where the numbers they use come from?

Any light you can shed would be most welcome

Cheers

Nobby

Mad (Flt) Scientist

10th Dec 2006, 16:43

There's something awry with that data; it's not internally consistent.

Ignore for the moment the pitch attitude and the fuel (I'm scratching my head over that; is this special fuel that has no weight or something?). Then the info reduces to:

An aircraft on wheels weighing

Aircraft main wheels are at 12.871M

Nose wheel 4.089M

Main jack points at 12.763

Nose jack 4.426

Though I think thats irrellevant (I agree, since it's specified as 'on wheels)

LEMAC 11.425

MAC 2.285

r = reaction on nose landing gear tyre

W = total wt

The H-arm M calc given is

12.908 - (8.832 r / W)

The XG(%MAC) given is

64.9 - (386.5 r/ W)

The H-arm equation is stating that the distance between the two weighing points - in this case the wheels - is 8.832m, and that the mainwheel weighing point is 12.908m aft of the reference. The data states that the values are (12.871m - 4.089m =) 8.782m and 12.871m respectively. That's errors of 5cm and 3.7cm respectively.

If you assume 10% of weight on nosewheels (a reasonable enough value) then the equation gives:

H-arm=12.908-0.1*8.832=12.0248m (excuse the excess decimals!)

while first principles gives:

H-arm=12.871-0.1*8.782=11.9928m

- a 3.2cm difference. For the quoted MAC that's almost a 1.5% discrepancy.

I'm already somewhat sceptical of the equations, but let's press on, assuming the one given is correct somehow.

Given that a=12.908 and b=8.832, previous posts have given:

c = 100/MAC*(a - LEMAC)

d = 100b/MAC

For LEMAC=11.425 and MAC=2.285

We should have:

c=64.9

d=386.5

At least the two equations given are consistent. Even if I'm not convinced either is correct.

Next, let's add the pitch attitude terms:

(The fuselage datum lime is at 3M if thats any use)

a = Aircraft pitch attitude (+nose up attitude)

The H-arm M calc given is

12.908 - 0.686 sin a - (8.832 r / W)

The XG(%MAC) given is

64.9 - 30 sin a - (386.5 r/ W)

As JT stated, the reason for the pitch attitude correction is the vertical position of the cg/centre of mass. If the mass all lay on one horizontal plane then you'd not need this correction, even at a pitch angle other than zero. Consider a ridiculous example of an aircraft with a huge AWACS-type mounting on its back, full of lead, such that the vertical position of the cg is as high above the reference as it is forward of the mainwheels. If we pitched the aircraft up to 45 degrees and looked at reactions we'd see there was NO reaction on the nosewheel, and with no correction we'd end up assuming that the cg is at the mainwheels - which it is, but only at this silly pitch attitude.

The pitch terms are making a similar correction, but for a more realistic case. They still heavily depend on ASSUMING the vertical cg position - and you may wish to consider how its possible to know the vertical cg if you don't know the weight or the longitudinal cg when you start the weighing .... (hint: it isn't - its an eductaed guess)

Looking at the correction term: - 0.686 sin a; we can see that it provides a smaller value for H-arm (i.e. moves the calculated cg position forward) if the aircraft is pitched up. This means that the raw, uncorrected, value is too far aft, which means that the vertical centre of mass is ABOVE the reference. The 0.686 tells us it's 0.686m above the reference.

In the MAC equation, the 0.686m becaomes 100*0.686/2.285=30.0, which it indeed does...

Since the fuel term is a correction to the 0.686m term, it follows that its an attempt to correct for vertical cg movement with fuel. It appears to be stating that the vertical cg moves by 0.00016m per kg - or, rather less annoyingly, 0.16m per tonne. Interestingly, this term is additive, which indicates that fuel is moving the cg upwards - which would seem to say it's either a high wing type with integral fuel, or fuel in the fuselage. Whatever it is, its another source of error, and best eliminated if you want a real answer.

PS thinking a bit, I wonder if the 5cm and 3.7cm apparent errors are due to either gear travel under load (trailing link type?) or some kind of wheel contact area offset.

nobbyknownowt

10th Dec 2006, 17:28

Wow!

thanks for your time and trouble. i think i can see what they are saying even if i cant see how they are saying it.

guess i am going to have to ask them direct how they get their numbers?

one would think, there is an explanation somewhere.

cheers

nobby

john_tullamarine

10th Dec 2006, 22:12

MFS obviously is having a slow day and has too much time on his hands ...

Clearly nobbyknownowt has some paperwork with this stuff on it .. how about scanning the sheets and emailing them to me and MFS .. at the moment the problem is a bit like trying to find something in a bag with a couple of sticks held in the hands ... while looking over one's shoulder using a mirror ....

If we had the original data, then there are likely to be some other clues to what the original derivation was ...