energy requirements

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Nicodemus38

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I am curious on how much energy is actually needed to operate the revolver, double action and single action by pulling directly on the hammer. and on the trigger for da.

i am curious on how much it differs between loaded and unloaded, and different standard weight ammo.
 
DA pull is probably around 8 to 12 lbs depending upon the revolver. Much depends upon the springs and how smooth the fit is in the moving parts. A colt Python is much less effort than a typical Taurus.

You'd have to have a much more sensitive finger to tell the difference in ammo by the DA pull than I have.
 
I am curious on how much energy is actually needed to operate the revolver, double action and single action by pulling directly on the hammer. and on the trigger for da.

Is this in regard to the amount of mechanical work done by one's hand, as in a physics question? If so, then it would vary between different revolvers--not just models, but individual instances--which requires measurements. To take a "typical" example, a revolver in DA mode might have a middling trigger weight of 10 pounds and a travel of 0.5 inches (in an arc, although it makes little difference in this case), which would work out to 10 * 0.5 / 12 = ~0.42 foot-pounds.

i am curious on how much it differs between loaded and unloaded, and different standard weight ammo.

Not much. :)
 
DA pull is probably around 8 to 12 lbs depending upon the revolver. Much depends upon the springs and how smooth the fit is in the moving parts. A colt Python is much less effort than a typical Taurus.

That's debatable gun to gun. I have a M85SSUL that compares well to any Python I've handled and I've fired several. On the average, you're right, though. But, Taurus triggers compare well to any current production revolver I've yet to pick up. I never handled a Korth, of course.

I'm not really getting the intent of the post. By "energy" are you meaning human effort to pull or are you talking about the actual kinetic energy of the hammer powered by the spring as it strikes the primer? DA revolvers require practice as regards to effort and marksmanship firing DA. Same with a DA autoloader, true DA or DAO. I don't know that human effort can be quantified and I don't know that the actual energy of the hammer at primer strike differs from what is required in an auto. I mean, it takes the same energy to cap the same primer. The effort that it takes to actually fire a DA weapon is as much dependent on the smoothness of the action as it is upon the actual weight of the trigger pull. I'll take a heavy/smooth trigger over a lighter/rough trigger any day.
 
recap, how many pounds of force are required to operate the action of a revolver, and how much does it change with weight of cartridges in cylinder?

ie, scientific experiment:

1. put revolver in vise
2. pull rearward upon the trigger with a poundage scale until DA mechanism makes the gun fire.
3. repeat for loaded state using multiple cartridge weights
4. repeat experiment by pulling rearward upon hammer only to cock the weapon to calculate the necesasry amount of force.
 
Your average street revolver will have about a 12lb+ pull. It'll take about that to cock the hammer with your thumb.

You would need a very accurate digital gauge (probably 5 or 6 decimals) to be able to measure the difference in the trigger pull between a loaded and unloaded revolver. AFAIK there is no such gauge, so you'll have to build one yourself.

Let us know what you find out.
 
A unloaded revolver is useless

While it may be a interesting question, I see no benefit in knowing the answer. A unloaded revolver is useless, so knowing its empty DA trigger pull weight is of little merit. Might as well ask what is the the difference in the SA hammer pull if its empty or loaded. Loaded with 38 spl. 148 grain HBWC or .357 magnums with 172 grain LSWC?
Much the same as being asked in 1973 what the weight of a M-60 MG is. Some one had to carry it regardless of the answer.
 
recap, how many pounds of force are required to operate the action of a revolver, and how much does it change with weight of cartridges in cylinder?

The pounds of force required remain the same; acceleration rate is reduced, however.

Think about it this way. Take a 2,000lb car and a 5,000lb car, and adjust tire pressures on each to equalize their stationary rolling resistance. Place them next to each other on a perfectly flat surface. You should be able to push either one with the exact same amount of force; the 2,000lb car will simply accelerate faster than the 5,000lb car, given the same force behind it.

In the same way, adding cartridges of equal mass to the cylinder will increase its overall mass and weight, but in a symmetrical fashion. No extra force is needed to turn it; it will simply turn slightly slower with the same amount of force.
 
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