Re: 7mm vs. .30 cal
Yes, Hired Gun has this correct.
We just happened to have this same discussion under the subject Stubby--Lapua in the general discussion forum, where we compared 7mm to 338 velocity. But the explanation is the same for the 7mm and 30 cal. Below I've copied and pasted the explanation from the Stubby--Lapua post.
"if equal pressures (say 60,000 psi) are generated in two different cartridges over an equal duration of time, the force exerted on the base of the bullet with the larger diameter will be greater than the force exerted on the base of the smaller diameter bullet. The area across the base of a 7mm bullet is about 0.0634 square inches. The area across the base of a .338 bullet is about 0.0897 square inches. The pressure generated by the cartridge acts against the base of each bullet. In the case of the 7mm bullet, the 60,000 psi creates a force of (60,000 lbs/square inch) X (0.0634 square inch) = <u>3804</u> lbs of force. With the .338 bullet the 60,000 psi pressure creates a force of (60,000 lbs/square inch) X (0.0897 square inch) = <u>5382</u> lbs of force.
We can see that even though the pressure is equal in each bore, much more force is applied against the base of the .338 bullet than the 7mm bullet, because the force exerted is a function of the area that the pressure is able to act on. And I think we can all agree that the greater the force acting on two bullets of equal weight, the faster the bullet will be accelerated down the bore. Which is why bullets of equal weight will always be driven faster in a larger caliber, provided that the same pressures are maintained over the same length of time. So this has a straightforward explanation."