.45 Colt mostly because it does the same thing as a .44 Magnum at less pressure, about 10kpsi less when both are loaded to the max.
One of the gun world's oft repeated old wives' tales. Newton's 3rd Law also has an issue with it.
Perhaps you don't understand Newton's third law. The 45 has 10% more area over which that pressure is acting, and hence at the same pressure as the 44, the 45 is going to have a greater reaction. Conversely, the 45 requires less pressure than the 44 to achieve the same velocity, a proper application of physics not some old wives tale.
The bearing surface of the 45 Colt is approximately 15% more than the bearing surface of the 44 mag, thus the Colt 45 is acting against more friction with the contact of the interior of the bore than the 44 mag. How much does that off-set the 10% larger surface area for pressure to act on of the rear of the 45 Colt bullet?
I'm curious where you come up with the greater bearing surface on the 45, looking at bullet designs on the Mountain Molds website I come up with less bearing surface for the.
For the same weight bullet the 45 is shorter and assuming one uses the same nose length i.e. forward of the canalure on an ogival wadcutter i.e. lfn shape the 44 has a longer base and longer driving bands than the 45 for a given weight. In the case of a 250 gr with a .35" nose and 75% meplat the 44 had roughly 10% more bearing surface due to the longer base bands for a single lube groove bullet.
If you want to have a closer comparison it's better to use a 10% heavier bullet and at that point the 45 has 10% more weight, base dia, wound channel and due to lower operating pressure about 100 fps slower all while burning the same amount of powder than the 44.
Everyone has their preferences, but the math doesn't lie.
