# How much does rider weight increase tire pressure?

When a person sits on a bike, the air pressure in the tires must go up because of the additional weight. With 100 PSI in 700x23 tires on a riderless bike, what is the air pressure when a 100 lb rider is sitting on the bike? A 200 lb rider?

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The "spot" the tire makes on the pavement must grow from essentially nothing to an area sufficient to counteract the weight of the rider. But the tire pressure does not need to increase much to accomplish this, especially at 100 psi. Exactly how much the pressure increases could probably be (approximately) calculated but the math would involve the volume of a torus, etc, stuff I wouldn't touch (especially since I'm on vacation for the next two weeks). –  Daniel R Hicks Apr 18 '14 at 22:24
Consider: If you had an infinitely large tire (or at least a very large one) pumped up to 50 psi, with 50 pounds of weight on the tire there will be 1 square inch of contact. Add another 50 pounds and the contact patch increases to 2 square inches. Has the pressure in the tire doubled? No, the increase in contact area "bears" all the added weight and the pressure doesn't go up at all. –  Daniel R Hicks Apr 19 '14 at 0:41
@DanielRHicks - Maybe I am missing something, but your reasoning seems flawed: you assert that the patch increases to 2 squared inches (because you use the 50 psi figure) and then you use the 2 squared inches to assert the pressure hasn't risen. I could argue with 100 pounds on the tire, the patch goes to 1,2 inches and the psi to 83.34, or any other combination of numbers... or am I missing a key element of your explanation? –  mac Apr 20 '14 at 22:33
@mac - It's an infinitely large tire. –  Daniel R Hicks Apr 20 '14 at 22:34
@DanielRHicks - Ok, now I get it, although I could argue that if I sit an infinitely heavy rider on it... ;) –  mac Apr 20 '14 at 23:40

An alternative way to calculate the solution:

The pressure in the tyre is 100psi. Pressure equals force divided by area. The surface area inside the inner tube (for a 700x23c tyre) is (very) roughly 7.2cm x 210cm = 1512cm square [or in square inches = 234insq]. The total forces involved on an unladen wheel are therefore 100 ‘pounds per square inch’ x 234 square inches = 23400 ‘pounds’. By sitting on the bike, one adds another 100 pounds to the wheel and thus to the forces, ie 23500 ‘pounds’. Assuming the inner tube does not change surface area ie the wall in contact with the ground is deformed in shape but overall not stretched or shortened, then the final pressure will be 23500 ’pounds’ divided by 234insq = 100.4 psi. Overall an increase in pressure by 0.4%.

I have used different assumptions/ approximations to previous posts:

1. The dimensions for my inner tube are actually external dimensions rather than internal.

2. The tyre (and therefore the inner tube) is assumed not to measurably stretch when at an appropriate pressure. Similarly the deformation of the tyre wall from curved to flat is assumed not to change the surface area much.

3. Overall designed to be relatively quick maths (and for the benefit of reading I simplified to fewer significant figures along the way) so still prone to a largish margin of error.
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This matches my June 5 comment above. –  xpda Jul 14 '14 at 18:35
@xpda: and it's also wrong, just like your comment. Even just looking at the ideal gas law (PV=nRT) you should be able to guess that "change in pressure is related to change in volume" and conclude that without a lot of mathematics proving it anything that talks only about area is likely to be wrong. Especially given an answer that uses volume to come up with a very different answer. –  Mσᶎ Aug 1 '14 at 6:18

As a first approximation you could use zero.

But more accurately, with a 23mm tyre you probably have about 20mm diameter of air inside. A 700c wheel is ISO 622, so has an inner radius of 311mm. So your tube forms a torus with major radius 321 and minor radius 10mm. A litre is a cubic decimetre (a cube 1/10th of a metre or 10 cm on a side) so it's easier to use dm for this. So, major radius = 3.21dm, minor = 0.1dm.

Volume of a toroid = 2 π² minor² major = 2 π² 0.1² 3.21 = 0.63 litres

Now, at 100psi with a 200lb rider the flattened area is 2 square inches, or 1290 mm². So we want the volume of a toroidal section having that flat area. Conveniently the section is an ellipse, so the area is π (major radius)(minor radius) (and a circle is πr² because both radii are the same). As a first approximation let's say that the minor axis is half the width of the tyre (so 10mm) and see what happens:

a = π R r => R = a / π r = 1290 / 10π = 41mm

That seems a little on the short side, but it's vaguely plausible, so what's the volume? Unfortunately the volume of a toroidal section is a bit beyond my rusty calculus, so I'm going to cheat. A lot.

First, what arc does this represent?

sine θ = 41/321 => θ = 0.0004 radians (about 7°)

That's about 3.5° each side of the centre of the flat spot.

How deep is it? Since sine X = x for small x, and we definitely have small x, if we go out 20.5mm from the centre of the flat spot we've got a right triangle 20.5mm on the long side. The short side is that times sine 3.5° as above:

depth = 20.5 * (20.5 / 321) = 1/321 = 0.003mm
= 0.00003dm (since we're using dm for volume calculations)

As you can see that's a pretty flat shape, so even a crude pyramidal approximation is not going to matter too much (if it's out by a factor of 5 or 10 it's not really going to matter, it's still going to be "roughly zero" as a fraction of the total volume above).

Conveniently we can express the volume of a pyramid as a factor of the area of the base:

volume of a pyramid = 1/3 height * area of base

The area from above was 1290 mm², which we need to divide by 100² to get it in dm² for the volume calculations = 0.129 dm²

volume lost = 1/3 * 0.00003 * 0.129 = 1.34e-6 litres

What fraction of the original volume is this?

fractional change = lost volume / total volume = 1.34e-6 / 0.63 = 2.12e-6

See, even if that's out by a factor of ten, we're still looking at a volume change of 10-5. So the pressure change will be about the same, a change of 1/1,000,000 the actual pressure.

If this was the XKCD guy's "What If" I'd now work out how fat you'd have to be to cause an appreciable change in tyre volume. But I'm not, so I shan't.

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Bravo! Of course the numbers are out by a factor of two, because only about half the weight is on each wheel. Half of nothing is still nothing :-) –  andy256 May 9 '14 at 0:28
@andy256 but it's a smaller nothing :) And the OP was about a 100lb rider as well as a 200lb on, so there are really two answers: 5e-7 and 1.06e-6 respectively. Those zeros are getting pretty darn small. –  Mσᶎ May 9 '14 at 0:41
You might be interested in my calculations for a similar answer - specifically you might want to try my formula for the volume of the flattened bit of tire. I think the conclusion will be very similar. –  Floris Aug 29 '14 at 3:31
@Floris I think you used cross sectional area as a proxy for volume? Which is more accurate for a "rectangular toroid" car tyre, but even for a bike tyre it's very close. And for the sizes of zero we're talking about, I agree that the difference is also close to zero :) –  Mσᶎ Aug 29 '14 at 4:35
@Mσᶎ - yes, that's exactly what I did. When you include the other dimension, you end up with a patch that becomes both longer and wider at the same time, and while the volume would therefore scale with $L^4$, the relationship with force (which was linear in $L$ in my analysis) becomes $\sqrt{L}$ ($A=L\cdot w$) so the final expression goes quadratic in $P$ (so $P^2$ instead of $P^3$). –  Floris Aug 29 '14 at 11:56

I'm sure it's possible to calculate this but my bike was only a few steps away so it seemed easier to just measure the pressure change. I connected a floor pump with a pressure gauge and pumped the back tire to 100 psi on a 700x23 tire. I then sat on the saddle and watched the pressure gauge. It didn't move. I stood up and tried a second time with the same result. I'm sure it changed a bit but it was less than the resolution of the basic pressure gauge on my pump. Or maybe it was just a lousy experimental design.

I weight about 180 pounds.

So, experimentally with two trials, the answer is the pressure doesn't change very much.

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No, the charts from Cole are not measuring pressure increase, but simply indicated "suggested" pressure for a given weight. –  Daniel R Hicks Apr 19 '14 at 0:35
I think it has to be accurate. I checked it myself (on a tire with a shrader valve) and got the same result -- about 1 psi increase from 45. –  xpda Apr 19 '14 at 0:53
No, this was an excellent experiment, and you're right that the problem is with the resolution of your pressure gauge. –  Mσᶎ Apr 19 '14 at 10:06

There's a very simple way to accurately estimate this: use Boyle's gas law which states that the product of pressure and volume is constant. Thus if the air volume in your tire decreases by 10% (which I think is a huge overestimate), then the new volume V2 is 0.9 times the old volume V1. Hence the new pressure P2 must be 1/0.9 = 1.11 times the old pressure, i.e. pressure increased by 11%. I'd conclude that pressure is barely affected by the weight of the person, if the tire has enough pressure to support you in the first place.

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Interesting way of looking at it. A road bike tire at 100 psi hardly deforms at all, so the pressure wouldn't change much. However an underinflated tire, would deform quite a bit. Of course, it would only deform until it meets the rim, in which case the rim is support you and not the tire pressure. For maximum change in pressure you'd want a large tire that wasn't sufficiently underinflated such that you could get maximum deformation. –  Kibbee Apr 19 '14 at 13:35

All the deformation where the tire is in contact with the ground is NOT space lost.
Only part of the deformation represents space lost.

If I push in a perfectly round balloon it will deform to a different shape. The volume in the balloon will go down and the pressure up. But the decrease in volume is far less then the volume displaced by my hand. Yes a balloon is elastic and will expand. This is true for even an inelastic balloon.

Under no load a tire is round. This is not by chance. The tire has a fixed amount of material. The tire will assume a shape that creates the largest volume for a fixed amount of material - a circle. Under the operating range we are examining the circumference of the tire does not change - the tire does not stretch.

Under load the tire deforms to a different shape - an ellipse.
It is not a perfect ellipse but pretty close.
Not all the area displaced by the contact area is lost.
For a fixed amount of material an ellipse does not create as much volume.
The space lost is circle minus ellipse.

The equation volume for the of a circle is:
π r²
Since the circumference (material) of the tire is constant should use circumference c
c²/4π
The equation for the volume of an ellipse is
π r1 r2
The circumference (perimeter) of an ellipse is complex so will just assume (r1 + r2)/2 = r

Assume under load the tire displaces to r2 = 2r1 at the point of maximum displacement.
Height reduced by 1/3 at the maximum point of displacement.
The ratio (at maximum point of displacement) is:
π r1 r2 / π r²
r1 r2 / r²
r1 * 2r1 / ((r1 + 2r1)/2)²
2 r1² / (3r1 / 2)²
2 r1² / r1² * (3/2)²
2 / (3/2)²
2 / (3² / 2²)
2 * 2² / 3²
8 / 9

So roughly a 10% loss of volume at the point of maximum displacement.
Over the length of the patch with ground a (nominal) average of 17 / 18.

For r2 = 2 r1 the length of the patch with the ground is about 2 r

On a 700 / 25

( 50*17/18 + (700π - 50) ) / 700π
(700π - 50 (1 - 17/18) ) / 700π
(700π - 50/18) / 700π
1 - (50 / (18 * 700π))
1 - .001263
.998738

Pressure increases inversely to volume
So a pressure increase of
1.001264729 00.13%

Lets look at a patch of 50mm X 25mm
5*2.5 / (2.54)² * 100 psi = 193 lb If I put 200 lbs on a 700 X 25 at 100 psi that patch size and height reduced by 1/3 seems about right.
A 200 pound force increasing pressure by 00.12% seems about right.

I realize I Moz looks at it from all the displacement is lost.
If only part of the displacement is lost then the my number should be smaller than his.
I did not review his math.
Not trying to prove him wrong - this is just how I look at it.

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You're right that there will be a displacement effect on the skin of the tyre that I've ignored. But then you seem to be assuming that cross sectional area changes in the same ratio as volume, which I suspect is not correct. If you started by working with volume of a toroid instead of area of a circle and repeat the maths I think you might get a better answer than mine. But also, I suspect, you'd modify my final number by less than 1%... and we'd be talking about really, really small values of zero :) –  Mσᶎ Jun 5 '14 at 22:26
@Mσᶎ But the starting point is a circle - a tire bearing no weight is circle. Not sure what you mean by cross sectional area changes in the same ratio as volume. I assumed a 1/3 height deformation based on observation. That also came out to reasonable contact patch the right area to support a 100lbs. If I had used 1/4 it would not have changed the number as the contact patch would still need to be the correct area. I am not tying to prove you wrong nor discredit your answer. Just looking at it from a different perspective. –  Frisbee Jun 23 '14 at 18:48
Down vote care to explain? Really what do you not agree with? –  Frisbee Aug 29 '14 at 15:20