reciprocating mass

[redhotguitar89]

I'm doing a report in my physics class on the physics of paintball. I haven't played much paintball, and I want to know if anyone can explain to me what reciprocating mass. I haven't been able to find much on the subject so any equations I could use in my report to explain what it is and how it works would be really helpful, thanks.

[neppo1345]

If i remember correctly from physics last year...reciprocating mass is anything of mass moving back and forth. The reciprocating mass in a mag would be the bolt moving back in forth...so just weigh your markers bolt and you have your mass... In a blow back marker (like a spyder), its the bolt and the hammer. (If you wanted to get fancy, you could factor in the weight of the paintball once the bolt contacts it in both guns.)
The lower the reciprocating mass, the less kick and more accurate the gun.

my $.02

[redhotguitar89]

Thanks for your reply, before I do my project I want to know if there is any equation I need to enter the mass of the bolt into, or if that mass equals the reciprocating mass of the gun. Also if you have any equation on how reciprocating mass will affect the accuracy of the paintball gun, which would really help, too. Thank you again.

[Dayspring]

Well, reciprocating mass can induce kick that will alter the accuracy of a gun. As for an equation? Don't bet on that. It's rather subjecctive. One person's idea of major kick is another's "not so bad" kick.

Reciprocating mass will be any moving part on the gun that is in line with the flight of the ball.

On a timmy, that would be the ram assembly, bolt, bolt pin.

Cocker- bolt, bolt pin, back block, hammer, lug.

Does this make sense?

[DaveSM]

Thanks for your reply, before I do my project I want to know if there is any equation I need to enter the mass of the bolt into, or if that mass equals the reciprocating mass of the gun. Also if you have any equation on how reciprocating mass will affect the accuracy of the paintball gun, which would really help, too. Thank you again.

The only way I could figure out would be to calculate the "moment de force" (corry but I only know the frech words for this) produced when the mass move forward at maximum acceleration. To do so you need a few equations. I will use a spyder as an example.

1st formula
F=k*(delta)l (f=Newton, k=Newton/meter and l=meter)
With this formula you'll ba able to calculate the maximum force produced by the spring but you'll need to know k (which is different from spring to spring). With this formula you'll be able to find the k of your spring. Then calculate the spring compression while at rest and also when fully cocked (you'll need a caliper to get accurate measurments). The you will have your l and k.

2nd formula
You need to know the "centroïde" of the mass you are analysing. The "centroïde" is the location in wich, if you calculate the effect of a force it would produce the same effect as the force produced by the whole mass. There is no easy way to do so... ( I usually calculate it on complex 2d geometry only...) Here is the way I would suggest.
Y'=(Y'1*m1=Y'2*m2)/(m1+m2)
Y'1 is the vercital location of the "centroïde" of the bolt (meter)
Y'2 is the vertical location of the "centroïde" of the striker (meter)
m1 is the mass of the bolt (kg)
m2 is the mass of the striker (kg)
I suggest you use the middle of each (bolt and striker) of these as their centroid as it would be a fair evaluation... You could use Y'1=0 and Y'2= the distance between the bolt's "centroïde" and the striker's "centroïde"

3rd formula
M=F*r (M=Newton*meter F=Newton r=meter)
F= the force you calculated above (in Newton)
r= the distance perpendicular to the force (so the vertical distance between the middle of your hand and the "centroïde") between the "centroïde" (Y') of the bolt/hammer assembly and the middle of your hand
You will then get the torque produced by the market each time it is shot. You hands must produce an equal but opposed torque to keep the marker steady.

I know you would have to consider that the mass is moving but at the beginning the hammer/bolt isn't moving so the full spring force is used to push it forward. Also this only take in consideration the forward torque but you could also calculate the backward torque (when the stiker is pushed back into position).

This won't give you an exact value but it will be a fair evaluation.

You could also use a single equation using all these.

M=(k*(delta)l)*((Y'1*m1=Y'2*m2)/(m1+m2)+D)

D is the distance between your coordinate system origin and the point representing the middle of your hands.

Good luck