[jeddedia]

I am trying to figure out the energy or force that is transfered to the bolt stop pin from the bolt when they contact after firing a round. Is there an easy way to do this or is there technical info somewhere on the web? Any info at all would help.:)

Thanks,
Chris

[Vern Humphrey]

Simple way (and close enough for government work) is the old equation M1 * V1 = M2 * V2.

M1 is the mass of the bullet and gas, V1 is the velocity of the bullet and gas (for absolute accuracy, you need to consider gas velocity at around 4,000 fps, with half the gas mass -- and calculate this separate from bullet mass and velocity.)

M2 is bolt mass. V2 (which is what you're solving for) is bolt velocity.

Mass and velocity of the gun enter the equation, but only marginally so -- since the bolt is at max velocity before the gun begins to move.

And remember that to get mass you must divide weight by the gravitational constant, 32.2 ft/sec/sec.

Once you have bolt velocity, calculate kinetic energy of the bolt by Ek=1/2 M*V^2.

[owen]

Vern got you as far as the energy. To get the force, you really need to know how long the two parts are stuck together.

m1v1-m1v2=Ft

Supposing the rifle, and therefore the pin is fixed (assuming the rifle can move starts to become a nightmare of variables). The change of momentum in the bolt equals the impulse imparted on the bolt by the bolt stop pin.

In this case, v1 is the bolt velocity before the bolt hits the pin, and v2 is the bolt velocity after it hits the pin. m1 is the mass of the bolt. F is the force imparted by the pin on the bolt, and t is the time of contact. please remember the velocity is a vector and not a scalar.

I highly recommend you use metric units for anything like this. It makes life much, much easier.

Of course, I'm sure someone will ccome along and cut this up. In my old age, this stuff is getting fuzzy.

[owen]

As an afterthought, I think you can use the Coefficient of Restitution to make a guess at the time of contact...

[jeddedia]

Thank you guys! Next stop: working for NASA:D

[hksw]

Must include the force of the recoil spring (comparitively small as compared to the recoiling force but it is not zero).