In the 1911 Winchester catalog recoil numbers are given for all of their cartridges. Black powder loads are listed at significantly higher recoil than smokeless with the same bullet and same velocity. How is this possible?

Because the amount of powder does significantly affect recoil calculation, even with the same resulting velocity and bullet weight.

A complete formula for recoil calculations includes the weight of powder with its own velocity, most often given as some 4700 fps.

Here's an example I pulled off thehighroad web site:

Free recoil energy (in f.p.e.) = ([MV + 4700 P] ÷ 7000)² ÷ (64.34809711) ÷ (W)

where:

M = weight of the projectile in grains
V = velocity of the projectile in feet per second
P = weight of the powder charge in grains
W = weight of the firearm in pounds

There are a few other considerations when comparing black powder to smokeless, but the above formula gives you the idea.

The increased weight of the powder. Of course! I can sleep at night now!

I would question the velocity of the powder. It would seem that it's velocity is restricted to the velocity of the bullet as long as the bullet is in the bore, and after the bullet leaves, it is 'water under the bridge'!

Even without a bullet, the compressed air leaving a balloon will fly it all around the room...

Yuh, likely. Thanks for the info.

Hmm, could it also be due to the fact that black powder is an explosive whereas smokeless powder is not. Smokeless burns very rapidly, and presumably progressively, thus the recoil is not instantaneous as in "explosion"?
Peter.

The in lies the reason that the British troops in the Peninsular campaign were always told to aim low when in line to the approaching French.Not just the combustion but the muzzle rise as well.