Going deeper on lesson 1 Q8. The lesson says contact-patch area = force / pressure. That is exact for a balloon and only approximately true for a tire, because a tire's walls carry some of the load. This is the model that says how much, the two limits, and the 20-minute experiment that measures it. Nothing here is a fact about our tire until that experiment is run and recorded.
1. Where the balloon formula comes from
Take a tire as a membrane with no stiffness. Inside, air at gauge pressure p pushes on every square inch. Where the tire is flattened against the ground over area A, the ground must push back with p*A, and that is the only thing holding the corner up:
F = p * A => A = F / pTire size, shape, and material don't appear. That is the surprising part: a big balloon and a small balloon at the same p and F make the same patch area, just a different shape.
2. Add a carcass: two springs in parallel
Real walls resist being squashed. Model the structure (the carcass: the rubber and cord walls) as a spring in parallel with the air. If the tire squashes by a deflection d, the carcass pushes back k_c * d, where k_c is the carcass stiffness in pounds per inch of squash, and the air pushes back p * A(d), where p is the gauge pressure and A(d) the patch area at that deflection:
F = p * A(d) + k_c * d (1)Geometry of the patch: flattening a cylinder of radius R by d makes a chord of half-length sqrt(2Rd); for a tire that flattens across its full tread width w,
A(d) = 2 w sqrt(2 R d) (2)(1) with (2) is a quadratic in sqrt(d). Solving it and then computing A tells you how far the real patch is from F/p. Our patch model (a short script) does exactly that; the table in section 4 is its output.
3. Where Young's modulus enters
k_c is not a material constant; it is a structural stiffness, and that is the honest place to stop for a tire. But you can see how E gets in. Treat the tire as a thin ring of wall thickness t, modulus E, radius R, width w, loaded across a diameter. Ring-compression stiffness scales as
k_c ~ C * E * t^3 * w / R^3 (bending-dominated)with C of order 1-10 depending on how the load spreads (Roark's formulas for a thin ring give C ~ 6.7 for a point load across the diameter, less for distributed contact). So:
- E up 2x -> k_c up 2x
- wall 2x thicker -> k_c up 8x (the cube is why sidewall construction matters far more than compound)
- tire 2x bigger -> k_c down 8x (big tires behave more like balloons at the same wall)
Membrane tension adds a second term, ~ E t / R, that matters once the wall is stretched by the inflation pressure; for kart pressures it's secondary to bending but not zero. Don't compute k_c from E; measure it (section 5). The scaling is what to take away.
4. The two limits
Define the dimensionless ratio (a pure number, no units)
rho = k_c * d_0 / F, d_0 = (F / (p * 2 w sqrt(2R)))^2where d_0 is the deflection the balloon model predicts, w is the tread width and R the tire radius from (2).
i.e. the load the carcass WOULD carry at the balloon deflection, as a fraction of the total. Then:
- rho -> 0 (very flexible wall, or high pressure): A -> F/p. Balloon. The sheet's formula is exact.
- rho ~ 0.1: patch about 4% smaller than F/p. Approximation fine.
- rho ~ 1: carcass and air share the load; patch ~ 25-35% under F/p.
- rho >> 1 (stiff wall, or low pressure): A -> (F/k_c)^{1/2} * const; patch no longer depends on pressure at all. That's a solid wheel.
Note that lowering p moves you toward the stiff limit: at low pressure the carcass's share grows, which is why "let air out for more patch" stops paying off, and why a nearly flat tire is stiffer than the formula predicts.
For our numbers (90 lb corner, 12 psi, R 5.45 in, w 5.5 in), the model gives:
k_c 0 lb/in -> 7.50 in^2 (100%)
k_c 100 -> 7.17 (96%)
k_c 400 -> 6.45 (86%)
k_c 1600 -> 4.99 (66%)At 8 psi the same k_c = 400 gives 76%: the balloon formula degrades as pressure drops.
5. What the model leaves out
- Patch width is not the full tread: the tread band has its own bending stiffness, so edges lift and the real patch is narrower and longer than (2). This pushes A below F/p independently of k_c.
- Contact pressure is not uniform: real footprints show higher pressure at the center (stiff tread) or at the shoulders (stiff belts), so the "average pressure = inflation pressure" idea is only average.
- Rubber is viscoelastic; k_c depends on temperature and on how fast the load is applied. A hot tire is a softer spring.
- Camber and cornering load reshape the patch. The static model is for a kart sitting still.
6. The experiment that makes it real (garage, 20 minutes)
Measure the actual patch and back out k_c.
- Kart on the ground, driver seated (or ballast to race weight). Bathroom scale under one rear wheel to read F for that corner.
- Slide a sheet of cardboard under the tire, chalk the tread, roll the kart onto the cardboard and off. Trace the print. Count the squares on grid paper, or measure length x width and take 0.8 x that for an oval. That is A, measured.
- Do it at 8, 10, 12, 14 psi cold. Four points.
- Compare each A to F/p. The ratio A/(F/p) vs p is the curve the model predicts; fit k_c from it (run the model at each p and adjust k_c until the numbers match).
If A/(F/p) is 0.9-1.0 at all four pressures, the balloon is good enough and lesson 1's Q8 stands. If it drops toward 0.7 at 8 psi, we've measured the carcass, and the answer to "how much patch does lowering pressure buy" is smaller than the sheet says. Either way, one photo of the chalk prints and the four numbers go in the kart's records, and this page gets a dated result.