Jump to content
Sign in to follow this  
bonked

[Help] Measuring the weight of a car

Recommended Posts

Alright, so today in my science class we attempted to measure the weight of my teacher's 2006 Honda Civic. He claimed that by calculating the surface area of each of the tires, and the PSI in each of the tires, we could find the weight of his car.

 

We used the formula P (Pressure)=F (Force)/A (Area) rearranged for this lab:

F=PA

 

Because we only had tire gauges that measured PSI and not Pa, we went with standard units (Inches for length/width, PSI for pressure, lbs for weight). The weight came out to be around 4000lbs, obviously WAY too high for a Honda Civic.

 

The posted weight was 2628lbs. However, the measurements we got could NEVER have worked out because the posted weight doesn't take into account gasoline. Also: the tread on the tire reduces the actual surface area (which means the surface area is wrong).

 

Does this concept even make sense? Is it possible to do this, and do my problems with the experiment make any sense? Is my science teacher dumber then I am?

Share this post


Link to post
Share on other sites

Does he carry a lot of radiation shielding? ;)

 

It seems to me that there are a couple of things that might screw up your results:

 

First, if you were to jack the car up so that none of the tires were touching the ground, the tires would still register a pressure. So what are you measuring? The air is compressed before you start: you have an initial pressure before you load the tires with the car's weight. Perhaps if you subtract the "unloaded" pressure from the pressure you read, your answer may be closer.

 

The second thing, which is probably minor, is that the tire sidewalls have some degree of stiffness, and probably account for a fraction of the support.

Share this post


Link to post
Share on other sites

Right, the surface area is wrong if you don't account for the tread. You might get an answer that's too big by maybe 40%, which means that the weight will be similarly off. So what happens if you reduce your surface area by a factor of 1.4? This is more an issue of GIGO (garbage in, garbage out) than faulty physics. If you know the data are bad, the physics analysis isn't going to transform it into a good answer.

 

The weight of the gasoline can be estimated and used to adjust your answer. Using physics, even, since density*volume gives the mass. It will be fairly small compared to the car's weight, since gas weighs around 8 lbs/gallon and you have at most 12-15 gallons in the car (you could estimate the amount by looking at the gas gauge), so it's around 100 lbs.


Merged post follows:

Consecutive posts merged

First, if you were to jack the car up so that none of the tires were touching the ground, the tires would still register a pressure. So what are you measuring? The air is compressed before you start: you have an initial pressure before you load the tires with the car's weight.

 

It's true there is some residual force in the tire and an unloaded pressure. But at the unloaded pressure, how much weight is it carrying? Zero. And the contact area is likewise zero. You could test this by seeing if the pressure changed significantly as the tires were loaded as you lowered the jack, or if the effect was simply the change in contact area as the load increased.

 

A more thorough experiment would be to do a measurement and then change the pressure in the tire, and do it again. Repeat for several values of pressure.

Edited by swansont
Consecutive posts merged.

Share this post


Link to post
Share on other sites

I knew this experiment was faulty at best. He said if we came within 20% PE (percent error) of the listed weight, we would get 2 extra points on our next test.

 

There was simply no way to get within 20% of the actual without fudging the numbers.

 

***

 

I was wondering how the PSI even helped if we didn't have an initial measurement, like you were saying. We need the inital pressure subtracted from the final which accounts for the car's weight. You also brought up a good point about the walls of the tires themselves.

 

Sounds like my teacher doesn't know what the heck he is talking about, and I usually stump him in class with questions he can't answer, however he should be able to answer them. He basically reads from the book.

Share this post


Link to post
Share on other sites
Right, the surface area is wrong if you don't account for the tread. You might get an answer that's too big by maybe 40%, which means that the weight will be similarly off. So what happens if you reduce your surface area by a factor of 1.4? This is more an issue of GIGO (garbage in, garbage out) than faulty physics. If you know the data are bad, the physics analysis isn't going to transform it into a good answer.

 

 

.

 

I think you have to be careful in assuming the error is related to the contact area of the tread. The gaps between treads are for the most part compensated for by "bridging", where there is an increased pressure locally where the tread contacts the road compared to the consistent pressure on the inside.

 

I think the error is more likely related to the area near the perimeter of the contact area where the contact pressure would be reduced compared to the inside pressure due to the stiffness of the tire. This reduced pressure area would be insignificant on something such as a balloon, where the stresses on the skin are almost purely tensile.

Share this post


Link to post
Share on other sites

The issue is clearly more complicated than F= PA.

 

If you could calculate the weight of the car solely by the air pressure and contact area, why is it that a tire normally filled to 40 psi will lift a car with only 10 psi? Clearly the surface area of the tire contact to the ground at 10 psi (while bigger) isn't 4 times the size when at 40 psi.

 

I think J.C. MacSwell is onto something...the rubber of the tire isn't simply in tension. There are tensile and compressive forces acting on the area in contact with the road and on the areas in contact with the rim.

 

If the experiment could be repeated, I'd try to measure the forces on the rubber directly. Perhaps you could measure the distance changes on the tire (at say 6 locations ) between the rim and the end of the tire by chalk marks and a string (as the tire deforms) before and after the load is applied. I'll bet you could determine the forces (either tensile or compression) on the tire at all the locations. I'd suspect (because tires with low air pressures are deformed) you'd get different forces at different locations within the rubber.

Share this post


Link to post
Share on other sites

Are you measuring the surface area of contact with the ground?

 

Also, since the tire is not perfectly flexible, that could mess with the numbers as well.

Share this post


Link to post
Share on other sites
I think you have to be careful in assuming the error is related to the contact area of the tread. The gaps between treads are for the most part compensated for by "bridging", where there is an increased pressure locally where the tread contacts the road compared to the consistent pressure on the inside.

 

I think the error is more likely related to the area near the perimeter of the contact area where the contact pressure would be reduced compared to the inside pressure due to the stiffness of the tire. This reduced pressure area would be insignificant on something such as a balloon, where the stresses on the skin are almost purely tensile.

 

Yeah, I was thinking about this after I posted, and the important number has to be the pressure on the wall of the tire, not on the tread — the pressure on the tread will not necessarily be the same as what is read on the tire pressure gauge.

Share this post


Link to post
Share on other sites

I clearly have sparked some thinking to be done on the topic. I agree that the problem is much more complicated the P=F/A and that there are numerous variables which can throw off the numbers.

 

How about this: apply paint to the tread of the tires. Drive the car over a piece of paper, creating an imprint. I will get my measurements from that piece of paper. Again, that only solves the tread's surface area problem. It doesn't fix the pressure problems.

Edited by bonked

Share this post


Link to post
Share on other sites
I clearly have sparked some thinking to be done on the topic. I agree that the problem is much more complicated the P=F/A and that there are numerous variables which can throw off the numbers.

 

How about this: apply paint to the tread of the tires. Drive the car over a piece of paper, creating an imprint. I will get my measurements from that piece of paper. Again, that only solves the tread's surface area problem. It doesn't fix the pressure problems.

 

You can check the pressure problem by doing the experiment at different pressures to see if it's a linear effect with a zero y-intercept or not. And also by comparing the unloaded pressure vs loaded pressure.

Share this post


Link to post
Share on other sites
I clearly have sparked some thinking to be done on the topic. I agree that the problem is much more complicated the P=F/A and that there are numerous variables which can throw off the numbers.

 

How about this: apply paint to the tread of the tires. Drive the car over a piece of paper, creating an imprint. I will get my measurements from that piece of paper. Again, that only solves the tread's surface area problem. It doesn't fix the pressure problems.

 

Instead, I'd paint the tires, jack the wheel off the ground, then set the tire onto a peice of paper. Then jack the wheel off the ground again and remove the paper. This will account for a change in shape in all directions, rather than just an increase in width of the tire. As such, you will have a better measurement of the tread surface area because there could be an increase in the length of the tires contact with the ground in addition to an increase in the width.

 

Ideally, this should be done for multiple air pressures. Perhaps you could measure it fully inflated, then let 5 or 10 psi of air out of the tire and measure again. Repeat until the tire was fully deflated. Make sure to deflate all tires at the same time to assure the weight of the car is being distributed evenly. (of course you should also re-inflate the tires when you are finished:D )

 

If you can show very little surface area change between 40 psi and 5 psi (i.e. the surface area at 5 psi isn't 8x the area at 40psi), then you should easily be able to disprove F=PA.

Share this post


Link to post
Share on other sites

Very good idea, SH3RLOCK.

 

I'd recommend using graph paper, as it might be easier to measure the area that way.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.