Weight of a plane

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  • #26
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Measure tire pressure before and after take off?
 
  • #27
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Measure tire pressure. Then take a pencial a draw the outline of the area of each tire in contact with the ground. make the plane move and calculate the area of each area you outline. Then the force exerted by the plane on the ground would be the pressure of each tire multiplied by its corresponding area of contant to the ground. Sum all three forces. Divide this by 9.81 and get the mass.
 
  • #28
use a screw driver from the toolbox to open hatch gr58 and look , using the torch at the pressure in the hydraulic undercarriage gauge then do the math to ge the total weight of the plane
 
  • #29
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Measure tire pressure. Then take a pencial a draw the outline of the area of each tire in contact with the ground. make the plane move and calculate the area of each area you outline. Then the force exerted by the plane on the ground would be the pressure of each tire multiplied by its corresponding area of contant to the ground. Sum all three forces. Divide this by 9.81 and get the mass.
This is the correct answer, all the rest are wrong.
 
  • #30
EnumaElish
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Measure tire pressure. Then take a pencial a draw the outline of the area of each tire in contact with the ground. make the plane move and calculate the area of each area you outline. Then the force exerted by the plane on the ground would be the pressure of each tire multiplied by its corresponding area of contant to the ground. Sum all three forces. Divide this by 9.81 and get the mass.
Don't you need to adjust for the stand-alone tire pressure (i.e., deadweight pressure)?
 
  • #31
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What is stand-alone tire pressure?
 
  • #32
you have not taken into account the elasticity of the ground

good try
 
  • #33
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What would that matter?
 
  • #34
see my tips for re-engineering
 
  • #35
just as the tyres will distort and take the load of the plane the ground will also to a degree do the same
 
  • #36
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But how does that affect the calculation? The area of contact has a uniform tire pressure acting on it. What the ground does makes no difference, all I care about is the contact area.
 
  • #37
Gokul43201
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Measure tire pressure.
Easier said than done! My tool box doesn't come with an airplane tire guage that goes up to several hundreds of psi.

Then take a pencial a draw the outline of the area of each tire in contact with the ground. make the plane move and calculate the area of each area you outline.
I don't imagine it's easy to "make the plane move", but that is not particularly important. A good estimate of the contact area can be made without moving the plane.

Then the force exerted by the plane on the ground would be the pressure of each tire multiplied by its corresponding area of contant to the ground. Sum all three forces.
There are six wheels, but that's just a detail.

Divide this by 9.81 and get the mass.
Sounds like a good plan! Only thing to figure out is how you'd measure the tire pressure. Remember, this is probably at least a couple hundred psi!

Now, if it were only possible to measure the contact area to an extremely high accuracy (better than 1ppm), you wouldn't need to know the tire pressure. Just measure the area twice, once with, and once without the toolbox on the plane.

If the tires had some kind of release valve, you could let the air out and measure its volume at STP (allow it to inflate a trash bag, or make it displace a liquid like airplane fuel). To improve the estime, I'd repeatedly measure the tire area as a function of volume of air released, plot it, fir it to an ideal gas curve and estimate the pressure. Alternatively, if the release valve has a cross section area smaller than a square inch, it may be easier to measure the force required to manually plug it.
 
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  • #38
if the ground is hard the contact patch will be less as the total mass of the craft will be exerted on the flat surface of the runway but if it is in soft ground the tyre will be sunk in and therefore the contact patch will be greater thus lowering the pressure per sq.in. as there will be greater contact on the surface area in contact
 
  • #39
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Airplane tire pressure is around 38-45 psi (for a small plane).

~200psi for a 747. (my road bicycle pump goes up that high, well almost 160psi max)
 
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  • #40
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if the ground is hard the contact patch will be less as the total mass of the craft will be exerted on the flat surface of the runway but if it is in soft ground the tyre will be sunk in and therefore the contact patch will be greater thus lowering the pressure per sq.in. as there will be greater contact on the surface area in contact
I agree. But I dont think the modulus of elasticitiy changes that much to cause a variation of more than a few hundred pounds. I think you would be pretty darn close using this method. Heavy aircraft are not rated to go on soft surfaces like grass.
 
  • #41
Gokul43201
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Airplane tire pressure is around 38-45 psi (for a small plane).

~200psi for a 747. (my road bicycle pump goes up that high, well almost 160psi max)
He he! I just googled up a picture of a 747, estimated the contact area and guessed the pressure would need to be about a couple hundred psi! My car tire gauge doesn't go up that high.
 
  • #42
Would all agree that two elements that may be crucial in all of this are . . . .
Altitude ( less gravitational effect and less atmospheric effect )
temperature ( creating expanded components thus lighter )

now watching Miami Ink Kat special
enjoy
 
  • #43
DaveC426913
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temperature ( creating expanded components thus lighter )
enjoy
Did you want to rethink this one? Perhaps explain how expansion causes items to become "lighter"? :rolleyes:
 
  • #44
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Ok, so here are the cold hard FACTS.

Being the dork I am, I went to the garage, got a couple sheets of thin newspaper, and slid them as far into the tire as they would go until they stopped. I then measured the distance between the two newspapers to find the lateral distance of the contact patch. I did the same thing to find the width of the patch. I then recorded all the tire pressures with a tire gauge.

Width: 6.625"
LF: 7" - 27.5 PSI
LR: 5.375" -27.5 PSI
RF: 6.75" -27.5 PSI
RR" 5.37" -26 PSI

These numbers seem very reasonable. Same pressure, so I get the same contact areas in the front two tires, and same can be said for the rear two tires. In addition, the front tires have more weight than the rear, GOOD its a FWD car.

Using my pocket calculator.

[tex]RF+LF+RR+LR=1299.765+1275.315+925.84+979.25= 4480.162lbs[/tex]

http://www.internetautoguide.com/car-specifications/09-int/1998/honda/accord/index.html

Curb weight (car + fuel) = 2888.05 pounds

My car was on an empty tank (so add the weight of 16 gallons of fuel too).

Conclusion: This looks nice on paper, but is total garbage in real life. This is because the tires have a stress distribution along the contact area that is not equal to the tire pressure. Sorry folks, the numbers dont lie. It over-estimated the curb weight by almost twice as much. 55.12% error :yuck:!
 
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  • #45
DaveC426913
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The final calculated weight will be very dependent on the actual area of contact.

Did you account for the fact that the contact area is not a rectangle? Your method for measuring it is very inaccurate. I would say the contact area is more like 2/3rds the area you measured.
 
  • #46
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Well, it was the easiest way. Just slide about 5 pages of thin newspaper until it wont go in any more. Then repeat from the other end and measure inbetween papers.

Well, of course its 2/3rds off!
 
  • #47
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i would go on google and search the model number and look up it's specs :D hehe
 
  • #48
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i would go on google and search the model number and look up it's specs :D hehe
Wrong.
 
  • #49
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The tyre won't put an even pressure on the ground over the whole contact area. Near to the edges of the contact area the pressure will be much less.

You can prove this for yourself using a car tyre. Get a thin piece of plastic, such as an old credit card - you'll find that it's easy to slide it under the tyre near the edges of the contact patch, but as you try to push it further in, it soon becomes impossible.
 
  • #50
Ceptimus is correct - to get Force we would have to integrate Pressure (points radially) dot Surface-Normal over the surface. As a true Theoretical Physicist, I am lazy, so let's make the work a little easier. First, change out the tires to square tires. Second ... OK joke

1. Let's assume all tires are equally weighted. Now I only have to get the force of one tire and multiply by the number of tires. Choose the baldest tire - the one that is most smooth, to reduce effects from tread.

2. Have the plane move so that this tire is on a hump, which I can create, reducing the contact area to something smaller, and curved down, opposite the up-curving tire. This will reduce the effects of area which has reduced force on it. Also, make the hump from something plastic, like clay, to fill in the tread. Now measure pressure and measure the area, multiply to get force.

3. As a good Theoretical Physicist, say that order of magnitude is good enough. ie anything from 1/5x to 5x is considered a success! Well, yes, the plane might crash, but I'll be able to estimate the impact force to an order of magnitude! And I'll have a tool box for my trouble!
 

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