Why do large volume wide wheelchair (or bicycle) tires roll over an obstruction more easily than small (narrow) tires? I have tested the force to pull a wheelchair over a $20$ mm high board and compared a 35mm wide tire and a $50$ mm wide tire.  I have adjusted for the larger wheel diameter in the $50$ mm tire since it has a mechanical advantage (in larger wheel diameter) going over the board.  I used a fish scale and a winch to smoothly pull the wheelchair over the board.  The tires are inflated to an equivalent firmness based on the casing tension formula.  The wide tire still rolls over the board with less force applied to the wheelchair.  Roughly $10.8$ kg for the wide tire and $11.8$ kg for the narrow one, and it is statistically significant in a t-test $p<.0001$.  Is it the compliance of a larger volume of air that would explain the difference?  Seems reasonable, but I have not come across how one would quantify that with an equation.
 A: I once spent an evening with a chap who owned a tyre manufacturing business and I recall that the physics of tyre design is much more complicated that I might have supposed without giving it much thought. However, that said I think the principles are probably along the following lines...
When you move an object from one height to another, the force required depends upon the direction of the force applied, the resulting direction of motion, and factors such as friction. For example, you require more force to lift an object directly upwards than you do to slide it up a lubricated ramp.
When you have a wheeled object meeting a vertical obstruction, the height of the obstruction relative to the diameter of the wheels is one factor that determines the force required. The other is the amount of give in the wheels. A pneumatic wheel will deform to some extent, the effect of which is equivalent to rounding off the corner of the obstruction, which makes the change in level less abrupt (ie more like movement up a ramp).
The question then is why does a balloon tyre give more than a narrow one. Going back to my evening with the tyre manufacturer, I am sure many factors might play a part, including the stiffness of the tyre walls, for example, but I think the overriding one is the ratio of the volume of the tyre to its outer radius. If you compare a low profile tyre with one with a smaller internal diameter, you will find that the volume of air in the former is less than in the latter. As a result, a given distortion in the circumference of the tyre leads to a higher relative change of volume and therefore a greater increases in pressure. The rise in internal pressure owing to the distortion of the shape of the tyre is what counteracts the distortion. In short, the balloon tyre will deform more because the resulting increases in internal pressure is less owing to its greater overall volume.
A: Assuming that the weight on the wheel is constant, the force resisting the pull can be determined entirely by the path it takes.
If you're pulling slowly forward, the moment of greatest required force occurs when the wheel hub starts to move in an upward direction, and the amount of force required depends on the angle of that motion -- steeper angle = larger force.  At that point, it's just like you were pulling the chair up a slippery ramp.
The angle at which the wheel hub starts to move upward is tangential to the wheel -- 90 degrees from the angle between the wheel hub and corner of the board, so the closer the wheel hub gets to being directly over the corner before it lifts, the less force will be required.
As @MarcoOcram details, the larger tire will deform more before it lifts, allowing the wheel hub to get closer to the board edge and therefore requiring a smaller pulling force.
