Checkboard,
Fully agree with your post and J P Davies. I've spent the last couple of days since this thread re-activated trying to internalise this. For a while there, I got lost in images of a military or business jet aircraft with small high pressure tyres slicing through standing water, while the big low pressure tyres of a widebody aquaplaned on top. That didn't help me internalise it very well.
Finally gave up, pulled out the books and built a spreadsheet to run some numbers. I used the L1101 and the BAC 1-11 (since I've got the data for both). Both run similar tyre pressures of 175/174 psi respectively (about 1200 kPa). At MTOW, the large main gear tyre on the L1101 has a footprint of 279 square inches, and on the BAC 1-11 the main gear tyre has a footprint of 136 square inches. Both have the calculated contact pressure of 175/174 psi (the same as the tyre pressure). Both should encounter dynamic aquaplaning at 119 knots and static aquaplaning at 102 knots.
Take the load off both planes, and the weight/tyre decreases. The radius of contact (and we assume here a circular contact area aka the footprint) decreases, and the footprint of the tyre decreases. Calculate the new ground contact pressure, and it remains the same at 174/175 psi. For the BAC 1-11 at 50,000 lbs weight, the contact area is down to 68.25 square inches, and the contract pressure is 174 psi. L1011 at 200,000 lbs has a footprint of 135.71 square inches and a contact pressure of 175 psi. In all the weight changes, the aquaplaning speed doesn't change.
One can intuitively visualise the onset of aquaplaning. The water is trying to push the tyre upwards off the runway, and this is being resisted by the weight on the tyre pushing down, together with the area (or footprint) of the tyre. These two are brought together as a single expression in terms of the contact pressure. The contact pressure is the same as the tyre pressure. So the aquaplaning speed is therefore represented by a function of the tyre pressure.