Find Weight of Passenger Jet Plane

[Cyrus]

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.

 

[Gokul43201]

Measure tire pressure.

Easier said than done! My tool box doesn't come with an airplane tire gauge 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.

 

[farpin1957]

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

 

[Cyrus]

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)

 

[Cyrus]

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 don't 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.

 

[Gokul43201]

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.

 

[farpin1957]

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

 

[DaveC426913]

temperature ( creating expanded components thus lighter )
enjoy

Did you want to rethink this one? Perhaps explain how expansion causes items to become "lighter"?

 

[Cyrus]

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.

$$RF+LF+RR+LR=1299.765+1275.315+925.84+979.25= 4480.162lbs$$

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 don't lie. It over-estimated the curb weight by almost twice as much. 55.12% error :yuck:!

 

[DaveC426913]

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.

 

[Cyrus]

Well, it was the easiest way. Just slide about 5 pages of thin newspaper until it won't go in any more. Then repeat from the other end and measure inbetween papers.

Well, of course its 2/3rds off!

 

[AFG34]

i would go on google and search the model number and look up it's specs :D hehe

 

[Cyrus]

i would go on google and search the model number and look up it's specs :D hehe

Wrong.

 

[ceptimus]

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.

 

[prof tipesh]

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!

 

[ahmedhassan72]

It is very simple i think even simpler than the solution i will suggest:(I think that this answer has been said and also the riddle is not accurate as the given clues is not clear)
-with a tool like a nail you can make a hole in the tire and measure the time air takes to get out
and get another tire (not installed to a plane) and make the same hole with the same tool and measure the time it takes to get out..(repeat the steps for all the plane tires)
so that when the pressure of the tire is x time is y and when pressure is a time is b you can get by that way the pressure acting by the wire and the mass of the aeroplane where P=mg/A where where you got P and you have g and can get A approximately you can get m
-Determine the engine force that the pilot will apply in his flight when the plane is in horizontal position then resolute the force of the engine to the direction of motion and against weight so the component against weight divided by 9.8 is the mass
-you can ask anyone from the airport who will really know
-by measuring momentum as it impinges with anything i put in it's way when it is starting to move so that the momentum of thing i put (block of wood) can be measured you know it's velocity and you can apply the law of conservation of momentum
-I think the solution is even simpler.

 

[drpizza]

Use the hammer to smash the plane into little pieces. Prop a ruled level on a screwdriver, and piece by piece, balance them with the hammer by adjusting the position of the screwdriver. If a piece is too heavy, smash it into smaller pieces. Use the formula weight * distance to hammer / distance to piece = weight of piece to find the weight of each piece. Find the sum of the weights of all those pieces.

Units will be in hammers.

 

[Poop-Loops]

I would compare it against something of known mass. For example, by ramming it into the said object I would see the effects. Something like a tall building would suffice.

 

[jaap de vries]

You know, if the toolbox has a torch, we could shine light from behind the plane and measure the deviation of flash light. (Hmm...I wonder if it's Flash light or Flash-light...).

I think it is 'flashlight'. No need to capitalize just because it emits photons.

 

[Einstienear]

oh...

There has been many questions and fierce arguments over whether or not an airplane will take off if on a conveyor belt "runway"...with fierce answers, the question has been deemed simple, but with complicated answers.

One of my simplest solutions to this was a thought experiment.

OK, we know the plane is meant to take off from the ground,(ground to air) and find it difficult to answer.

A simpler tackle at this would be to imagine the plane coming from the air and landing on the conveyor strip.

we know if the plane (from ground to air) would not take off because of the conveyor belt, it would stop in a quicker amount of time, we know that the plane would not stop quicker if landing on it , concluding that it would take off, if it was on a conveyor belt.

Einstienear

 

[Cyrus]

Dude, this has been answered years ago. Search it. Were NOT discussing this lame question anymore. (It does take off). Have a nice day.

The reason this keeps coming up in other forums is becuase those other forums are full of guys who don't know a damn thing about physics and thus talk out their exhaust stacks on answers using examples that don't mean jack squat.


Am I a little sick of seeing this question pop up? -Just a little.