Changing Point of Impact via Bullet Weight - Page 2

Old Grump · April 10, 2011, 11:55 AM

Quote:

Old Grumps DW does what I’d expect with 125’s and not with 158’s. Exploring the details of the load might reveal it. A .38 Special case with some 158’s doesn’t have a lot of powder space left, so barrel/cylinder gap bleed-off can become more significant with some powders. But I’m just talking off the top of my head there. I’ll have to take my own DW with me next time out and see if I can replicate that behavior?

Funny we were on this topic yesterday because after I signed off I went out and did a little draw and shoot practice with 4 of my guns. This target was my Dan Wesson model 15 with 4" barrel and fixed sights. I used Remington UMC 130 gr 38 spcl for these results
[IMG]

[/IMG]
15 yards one shot at a time with each draw. Winter rust is showing but if I had used my 158 grain loads I would have been clean over the top of the target or at the very least nicking the top of the paper since this was 15 yards going by past experience. Next time I will do a 25/50/100 yard comparison with both weights using magnums.

jbrown13 · April 10, 2011, 08:48 PM

Unclenick, I would like to pick your brain some more, if I may. What is the resultant force from momentum of a 158 grain bullet accelerated to 800 fps in 750 milliseconds? What is the formula you use to make this calculation?

I tried to research this, but what I found isn't making sense to me. I think I'm getting confused with the fps/s units, because what I'm coming up with seems like a lot of force. I'm just being curious in my senior years.

g.willikers · April 11, 2011, 01:15 PM

You stated load is 224 ft-lbs of energy.
If you are lazy like me just use this:
http://www.handloads.com/calc/quick.asp

jbrown13 · April 11, 2011, 01:29 PM

g.willikers, thanks for that link. I was more interested in the Momentum, which this link calculates at 18. Unfortunately, the link doesn't give a dimension to the momentum, but I'm "guessing" it is in ft-lbs. Doesn't seem like much. Couldn't a 200 lb. man resist any muzzle rise from the resultant force?

**Unclenick** · April 11, 2011, 02:27 PM

750 milliseconds (0.75 seconds) is an eon in gun time. I expect you mean microseconds? I'll assume you do.

In general, the English system of units is messy because you have to distinguish pounds force from pounds mass (lb mass = slug / 32.174…). It's usually easier to convert to metric, solve it, then convert back.

158 gr = 0.01024 kg
800 fps = 244 m/s
750 μs = 0.00075 s

The physics formula always has t0 and t1 and v0 and v1 and you use the difference in velocities over the difference in times to get acceleration, but in the case of the gun, you always start at 0 fps in the chamber, so you can just use the muzzle velocity and divide it by the barrel time to get acceleration, then multiply acceleration times mass to get force.

244 m / s / 0.00075 s = 325,330 m / s² = a = acceleration

From Newton:

F = ma

so, F = (325,330 m/s) × (0.01024 kg) = 3,329 N

1 N = 0.225 lbf (pounds force with the f added for clarity)

so 3,329 N = 748.3 lbf

Mind you, that number is just the average acceleration force over the length of the barrel. Actual force will vary in proportion to the pressure at any given point in the curve, being perhaps twice that near the breech when the pressure peaks, and half that near the muzzle when the pressure has fallen off. The ratio depends on the length of the barrel. A 3" revolver barrel with a fast powder might come out 3:1, a longer barrel will be more.

Momentum is just mass times velocity, or mV. In the English system it is in pound feet per second, or lb-ft/s. That's pounds mass, not pounds force, as you have in the ft-lb of kinetic energy (sometimes written ft-lbf because of that; others prefer to keep the potential confusion alive out of a sense of tradition). In the metric system, where there is no confusion, momentum is in kg-m/s.

1 lb-ft/s = 0.138 kg-m/s*

Your 158 gr bullet at 800 ft/s has almost 18.1 lb-ft/s of momentum, or 2.5 kg-m/s or 2.5 N-s.

*Because N = kg-m/s², kg-m/s can also be written as N-s, but because of the algebraic equality it's six of one, half dozen of the other.

jbrown13 · April 11, 2011, 05:04 PM

Unclenick, thanks for the reply.

I didn't realize I was this dumb. I'm going to quit thinking about what goes on internally and just shoot at the target. All this thinking is mind-numbing. I'm getting a headache.

PS - Is physics your vocation or avocation?

**Unclenick** · April 11, 2011, 05:16 PM

Well, you did ask. I recommend taking two aspirin and avoiding bright lights and loud noises for a couple of days, though somehow gunfire doesn't seem to count in that second category. It's soothing.

I'm an engineer, not a physicist, but it's an extension of all those geek high school years, nonetheless.

April 11, 2011, 01:15 PM	#28
g.willikers Senior Member Join Date: September 28, 2008 Posts: 10,442	You stated load is 224 ft-lbs of energy. If you are lazy like me just use this: http://www.handloads.com/calc/quick.asp __________________ Walt Kelly, alias Pogo, sez: “Don't take life so serious, son, it ain't nohow permanent.”

April 11, 2011, 02:27 PM	#30
Unclenick Staff Join Date: March 4, 2005 Location: Ohio Posts: 21,063	750 milliseconds (0.75 seconds) is an eon in gun time. I expect you mean microseconds? I'll assume you do. In general, the English system of units is messy because you have to distinguish pounds force from pounds mass (lb mass = slug / 32.174…). It's usually easier to convert to metric, solve it, then convert back. 158 gr = 0.01024 kg 800 fps = 244 m/s 750 μs = 0.00075 s The physics formula always has t0 and t1 and v0 and v1 and you use the difference in velocities over the difference in times to get acceleration, but in the case of the gun, you always start at 0 fps in the chamber, so you can just use the muzzle velocity and divide it by the barrel time to get acceleration, then multiply acceleration times mass to get force. 244 m / s / 0.00075 s = 325,330 m / s² = a = acceleration From Newton: F = ma so, F = (325,330 m/s) × (0.01024 kg) = 3,329 N 1 N = 0.225 lbf (pounds force with the f added for clarity) so 3,329 N = 748.3 lbf Mind you, that number is just the average acceleration force over the length of the barrel. Actual force will vary in proportion to the pressure at any given point in the curve, being perhaps twice that near the breech when the pressure peaks, and half that near the muzzle when the pressure has fallen off. The ratio depends on the length of the barrel. A 3" revolver barrel with a fast powder might come out 3:1, a longer barrel will be more. Momentum is just mass times velocity, or mV. In the English system it is in pound feet per second, or lb-ft/s. That's pounds mass, not pounds force, as you have in the ft-lb of kinetic energy (sometimes written ft-lbf because of that; others prefer to keep the potential confusion alive out of a sense of tradition). In the metric system, where there is no confusion, momentum is in kg-m/s. 1 lb-ft/s = 0.138 kg-m/s* Your 158 gr bullet at 800 ft/s has almost 18.1 lb-ft/s of momentum, or 2.5 kg-m/s or 2.5 N-s. Because N = kg-m/s², kg-m/s can also be written as N-s, but because of the algebraic equality it's six of one, half dozen of the other. __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member and Golden Eagle Last edited by Unclenick; April 11, 2011 at 05:11 PM. Reason: typo fixes and clarification*

April 11, 2011, 05:04 PM	#31
jbrown13 Senior Member Join Date: December 28, 2009 Location: Perry, GA Posts: 105	Unclenick, thanks for the reply. I didn't realize I was this dumb. I'm going to quit thinking about what goes on internally and just shoot at the target. All this thinking is mind-numbing. I'm getting a headache. PS - Is physics your vocation or avocation? Last edited by jbrown13; April 11, 2011 at 05:08 PM. Reason: added PS

April 11, 2011, 05:16 PM	#32
Unclenick Staff Join Date: March 4, 2005 Location: Ohio Posts: 21,063	Well, you did ask. I recommend taking two aspirin and avoiding bright lights and loud noises for a couple of days, though somehow gunfire doesn't seem to count in that second category. It's soothing. I'm an engineer, not a physicist, but it's an extension of all those geek high school years, nonetheless. __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member and Golden Eagle

April 10, 2011, 08:48 PM	#27
jbrown13 Senior Member Join Date: December 28, 2009 Location: Perry, GA Posts: 105	Unclenick, I would like to pick your brain some more, if I may. What is the resultant force from momentum of a 158 grain bullet accelerated to 800 fps in 750 milliseconds? What is the formula you use to make this calculation? I tried to research this, but what I found isn't making sense to me. I think I'm getting confused with the fps/s units, because what I'm coming up with seems like a lot of force. I'm just being curious in my senior years.

April 11, 2011, 01:29 PM	#29
jbrown13 Senior Member Join Date: December 28, 2009 Location: Perry, GA Posts: 105	g.willikers, thanks for that link. I was more interested in the Momentum, which this link calculates at 18. Unfortunately, the link doesn't give a dimension to the momentum, but I'm "guessing" it is in ft-lbs. Doesn't seem like much. Couldn't a 200 lb. man resist any muzzle rise from the resultant force?