Hello Alan,

rather simple, You just need to do a sum of the forces

at the bobsleigh.

- force driving downhill  = Mass x acceleration by gravity x sin angle of inlination.

You see, this force is much higher for a 4-bob, prop to mass. 

- friction between sleigh and ice = Mass  x acceleration (gravity) x coeffciant of friction.

This too is proportional to mass.

- air drag ("resistance") = about cross section x velocity squared.

You see, this one is the same for all bobs approximately.

The net force is dowmhill - frictoin - air drag = "rest"

@ george maybe the force pulling it downhill will be greater but it will be pretty exactly cancelled out by the  increase in inertia from the increase in mass causing no increase in acceleration.

When the weight is doubled, the the air resistance doesn't double.

The frontal cross section remains the same. The longer length would not double the drag as it would if they sat side by side.

Also when you double the weight , for example if you pick another person up while on ice skates, the drag of the skates on the ice might not double. I suspect it's  less than double.

The difference here is not in the simple physics of it, it's in the details of how drag and friction scale up, when doubling the number of men. Experience shows that these factors do not double.

i think part of the friction of ice is that pressure on it melts the ice creating lube therefore more pressure -more lube?