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?
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
  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 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.