Check weight of 747
 

Can someone tell me how you check the weight of a 747 after manufactured? Not using scales. This was a question I was asked at a interview...Help..riding4theson..thank you:confused:

I suppose you could measure the tyre pressure of every wheel and multiply it by the related ground contact patch area of each wheel and sum them all.

i've come across that question before. the answer i was told at the time had to do with putting it on a boat and then measuring the increase in displacement!! (hmmmm - any sane person is obviously going to think of that!) In theory, you could do an accurate calculation of weight, since the shape of the hull is known and the displacement would be large enough to be measured with some precision, however if i had a yacht large enough to float a 747 i'd stock up the bar and sail home from seattle, rather than fly my dubiously weighed jumbo...

johns idea is a good one, although I guess the internal stresses in the tyre structure will support a (probably very small) proportion of the weight, which would not show up in the tire footprint times pressure calculation.

kk

No .. the tyre footprint is a bit conceptual but, if one could do the relevant integration, the answer would be very close to precisely correct. You can't have some of the weight magically disappear ... although, following the fanciful boat solution, one could allow just a teensy bit of weight to be balanced by atmospheric hydrostatic forces integrated over the hull ... a la airships ... and, even then, one would need to consider the air inside the ship and whether that had any difference in pressure/mass to the air outside at the same RLs.

However, given that the normal weighing process is bedevilled by a range of errors ... we probably can ignore such minor considerations ...

The tire footprint technique may not work very well when you have multiple tires on one axle. It does work very well with cars. Try it yourself.

obviously the weight hasn’t magically disappeared. but the effect i mentioned will distort the calculation of it.
it depends what sort of answer you are looking for here. the tire contact pressure isn’t uniform, due to internal stresses and edge effects in the tire structure. you can integrate the pressure over the contact area, but first you have to know the distribution, which you don’t. this means any result would be approximate to some degree, depending how well you model it.

the boat solution is fanciful, but you allows you to compensate for every physical effect in your calculation. so in theory, you could calculate the EXACT weight, which was the objective of the question when it was put to me. the point is that, as a 'thought experiment', this gives the answer precisely, to arbitrary accuracy, the other method cannot.

cheers, kk

agreed .. but did you ever find out what they did have in mind with the question ?

I dunno, I thought I might fly it until it stalls, then record the CAS and compare that with the DPI figures!

Whilst also taking into account the density altitude at the time!

Might work, it would be at least as accurate as those electronic scales they use today!

FD Cheers :uhoh:

Recent electronic scales are pretty good. I have two sets of aircraft weighing kit which calibrate to around a kilo (two tonne) and 2-5 kilo (30 tonne) errors.

Mind you, some of the older kits are a bit suspect. All are VERY susceptible to inappropriate use during weighings .. lots of traps for young players ..

One idea may be to measure the oleo extensions on the undercarriage and work from there. In some AN2s I have seen they have marks on the oleo surrounds with weights. - I guess that it assumes they were at the correct pressure to begin with.:ok:

And CG X,Y & Z?

Why would you need to know the size of the contact patch to use the tyre pressure method? Wouldn't tyre pressure be proportional to the load on the undercarriage (compensating for temp. effects).

Pressure at a given starting weight is 'x'. Add a kilo & pressure is f(x+1). Add 2 kilos & pressure is f(x+2). Add 'n' kilos & pressure is f(x+n).

The relationship could be previously determined using an undercarriage set in a jig.

Tinstaafl,

You're correct if the volume of the tire changed considerably, and if you could measure the tire pressure when they were unloaded as well as when they are loaded. However, neither is true. What you do is multiply the pressure by the area that is supporting that pressure and you get the force due to the weight of the 747.

The reason that the volume change is considered negligible is because the pressure and volume can stay mostly the same, but the shape/size of the tire contact area changes. Of course, we don't really care much how the pressure changes because we only meter the laden tire.

The change in pressure is essentially what is happening with the oleos, but with the oleos, you do need to unload them first, so not as practical to do with everyday objects.

Going on from SCruisers point, adjust the pressure in the oleo by a known amount. Measure the increase/decrease in movement, then calculate the load.

load cells
 

I thought that it was done by strain gauges or load cells as in the Davco "Jack Weigh" system.

http://www.davco.bc.ca/jackweigh.html

The technique is surely to measure the contraction of a calibrated cell.