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9/22/2017 12:11:25 AM
Posted: 12/22/2005 8:19:53 AM EDT
[Last Edit: 12/22/2005 11:43:56 AM EDT by Pthfndr]
Ballistic calculators give a bullets energy at X distance in ft-lbs.

Is that the same thing as "Force"?

Ex: A 175 SMK with a muzzle velocity of 2700 fps has approximately 588 ft-lbs of energy at 1000 yards.*

Does that mean the the "force" of the hit is the same as getting hit with 588 lbs by an object .308" in diameter? What would the "force" be if the projectile was .264" dia with 543 ft-lbs of energy?*

*Figure calculated using the JBM online ballistics calculator. His page on energy:
www.eskimo.com/~jbm/equations/energy.html



Edited to correct ft/lbs to ft-lbs to make Mike Mills happy
Link Posted: 12/22/2005 8:25:11 AM EDT
No, this is the kinetic energy of the bullet equal to the potential energy stored in a 588lb weight lifted one foot off the ground.
Link Posted: 12/22/2005 8:28:16 AM EDT
It's confusing, too, because a common measurement for torque is pound-feet, especially when discussing engine output.
Link Posted: 12/22/2005 8:30:37 AM EDT
Force = Mass X Acceleration

Kinetic energy = (Mass x Velocity squared)/2
Link Posted: 12/22/2005 8:31:56 AM EDT

a common measurement for torque is pound-feet, especially when discussing engine
output.



Torque is rotational energy around a pivot point.
It's also referred to as a "moment arm".
Link Posted: 12/22/2005 8:43:25 AM EDT

Originally Posted By TUMOR:
Kinetic energy = (Mass x Velocity squared)/2



ahhh... ARFCOM

KE= 1/2 x m x v^2, definitely NOT the same as (mv)^2
Link Posted: 12/22/2005 8:49:43 AM EDT

Originally Posted By TUMOR:
Force = Mass X Acceleration

Kinetic energy = (Mass x Velocity squared)/2



Care to translate that into a number or something a layman can understand?
Link Posted: 12/22/2005 8:58:03 AM EDT

Originally Posted By TUMOR:

a common measurement for torque is pound-feet, especially when discussing engine
output.



Torque is rotational energy around a pivot point.
It's also referred to as a "moment arm".



Torque is NOT energy, it represents FORCE. The units appear the same but they are not as energy is s scalar quantity and torque is a vector. If you torque a bolt to 30 lb feet, that means you exert 30 lbs at the end of your ratchet if it is 1 foot long, 60 lbs if it is .5 ft long, etc.. This has nothing to do with energy since it does not factor in distance or velocity.
Link Posted: 12/22/2005 9:01:08 AM EDT
Tag for when teh fizz comes in to set things straight.
Link Posted: 12/22/2005 9:10:40 AM EDT

Originally Posted By Pthfndr:
Ballistic calculators give a bullets energy at X distance in ft/lbs.

Is that the same thing as "Force"?

Ex: A 175 SMK with a muzzle velocity of 2700 fps has approximately 588 ft/lbs of energy at 1000 yards.*

Does that mean the the "force" of the hit is the same as getting hit with 588 lbs by an object .308" in diameter? What would the "force" be if the projectile was .264" dia with 543 ft/lbs of energy?*

*Figure calculated using the JBM online ballistics calculator. His page on energy:
www.eskimo.com/~jbm/equations/energy.html





The energy can be identical for all sorts of combinations of bullet speeds and weights, or even other objects.

The terminal effect of bullets has as much to do with its momentum and diameter, and to complicate things, the bullets design (fmj, solid copper, solid lead, jhp, and so on). For example, a high strength solid bullet could be designed that would strike the target with the same energy as a pure lead bullet, but the high strength bullet would pass through both sides without changing shape or imparting much damage (it would because I designed it to) and the lead bullet would upset slightly and stay inside the target, implying that it shed 100% of its terminal energy in the target.

Bullet energy doesn't tell the whole story about its performance, but those numbers are usually thrown around the most when comparing two cartridges or bullets in the same cartridge.

Back to your original question, here are two examples of forces: your weight and how "hard" you can push on an object. Energy, work, and torque all share the same dimensions, i.e., foot-pounds, Newton-mm, inch-ounces, and so on, but they are not precisely the same physical phenomena (although they are related in an esoteric, philosophic way), and the context of their usage has to be understood.
Link Posted: 12/22/2005 9:45:49 AM EDT
another layman's explanation:

force = a push or pull
work = a push or pull applied over a distance (pushing on your house will make you tired, but you'll accomplish zero Joules of work)
energy = a capacity to do work (also measured in Joules)

someday the metric system will win.
Link Posted: 12/22/2005 9:54:56 AM EDT
[Last Edit: 12/22/2005 9:55:11 AM EDT by TexasEd]

Originally Posted By DukeSnookems:

Originally Posted By TUMOR:
Kinetic energy = (Mass x Velocity squared)/2



ahhh... ARFCOM

KE= 1/2 x m x v^2, definitely NOT the same as (mv)^2



Looks the same to me. He has mass times the velocity squared (not MV squared) over 2

Link Posted: 12/22/2005 10:03:37 AM EDT

Originally Posted By TexasEd:

Originally Posted By DukeSnookems:

Originally Posted By TUMOR:
Kinetic energy = (Mass x Velocity squared)/2



ahhh... ARFCOM

KE= 1/2 x m x v^2, definitely NOT the same as (mv)^2



Looks the same to me. He has mass times the velocity squared (not MV squared) over 2




Ahh, you guys are right, registered it wrong, no more posting right after waking up for me.
Link Posted: 12/22/2005 10:59:28 AM EDT
[Last Edit: 12/22/2005 11:01:26 AM EDT by Pthfndr]

Originally Posted By AeroE:

The energy can be identical for all sorts of combinations of bullet speeds and weights, or even other objects.

The terminal effect of bullets has as much to do with its momentum and diameter, and to complicate things, the bullets design (fmj, solid copper, solid lead, jhp, and so on). For example, a high strength solid bullet could be designed that would strike the target with the same energy as a pure lead bullet, but the high strength bullet would pass through both sides without changing shape or imparting much damage (it would because I designed it to) and the lead bullet would upset slightly and stay inside the target, implying that it shed 100% of its terminal energy in the target.

Bullet energy doesn't tell the whole story about its performance, but those numbers are usually thrown around the most when comparing two cartridges or bullets in the same cartridge.

Back to your original question, here are two examples of forces: your weight and how "hard" you can push on an object. Energy, work, and torque all share the same dimensions, i.e., foot-pounds, Newton-mm, inch-ounces, and so on, but they are not precisely the same physical phenomena (although they are related in an esoteric, philosophic way), and the context of their usage has to be understood.



Ok, I grasp all that. The reason I ask the original question is because over on the precision rifle forum someone asked about the 6.5x55 (.264) round as a long range sniper round. Which got me thinking. I shoot metallic silhouette competition. The ram silhouette we use is made of steel plate and weighs about 65 lbs. It's not unusual for people to have "ringers", where the bullet makes a good hit but the ram does not go down. Because the ram is steel I would think that the bullet sheds 100% of it's "energy".

I've seen the .264 bullet, going with sufficient velocity knock the ram down, and the .308 bullet - heavier but going slower - fail to knock it down.

No doubt the bullet design, and what it is hitting determines the terminal effect. Real world testing will reveal what that would be, whether on steel or ballistic gel.

But, assuming a bullet is going x fps and shedding 100% of it's energy, is it possible to give a number that tells what force ( such as lbs per square inch) it is hitting an object with? Feel free to use one of the examples in my original post.
Link Posted: 12/22/2005 11:05:06 AM EDT
From what I know, it's more complex than calculating the force the bullet strikes the steel plate.

It's just like a bowling ball hitting pins, a baseball bat smacking a curveball, etc.

It's more complicated than a mere measurement of force, and involves both a momentum transfer as well as energy.

If the bullet ricochets off the target, it is not the same as a purely elastic collision in which all of the energy is transferred as kinetic. It might be close, but isn't 100% elastic.

Somebody with a physics background can end all of this. I studied raccoon tracks in college.
Link Posted: 12/22/2005 11:13:38 AM EDT
[Last Edit: 12/22/2005 11:14:00 AM EDT by Mike_Mills]
It is possible to do such a thing, Pthfndr, but without some real world data it might be hard to believe the answer. For instance:

impulse into target = momentum of bullet

Force on target * duration of impact = mass of bullet * velocity of bullet

Rearranging terms yields:

F = m*v/t

The duration of the collision affects the force. Bullet type (construction, materials,...) and possibly even the actual velocity versus time profile during the collision will affect the results.

P.S. - please, guys, at least get the units correct. It's foot-pounds not foot/pounds, they are different.
Link Posted: 12/22/2005 11:51:30 AM EDT

Originally Posted By Pthfndr:

Originally Posted By AeroE:

The energy can be identical for all sorts of combinations of bullet speeds and weights, or even other objects.

The terminal effect of bullets has as much to do with its momentum and diameter, and to complicate things, the bullets design (fmj, solid copper, solid lead, jhp, and so on). For example, a high strength solid bullet could be designed that would strike the target with the same energy as a pure lead bullet, but the high strength bullet would pass through both sides without changing shape or imparting much damage (it would because I designed it to) and the lead bullet would upset slightly and stay inside the target, implying that it shed 100% of its terminal energy in the target.

Bullet energy doesn't tell the whole story about its performance, but those numbers are usually thrown around the most when comparing two cartridges or bullets in the same cartridge.

Back to your original question, here are two examples of forces: your weight and how "hard" you can push on an object. Energy, work, and torque all share the same dimensions, i.e., foot-pounds, Newton-mm, inch-ounces, and so on, but they are not precisely the same physical phenomena (although they are related in an esoteric, philosophic way), and the context of their usage has to be understood.



Ok, I grasp all that. The reason I ask the original question is because over on the precision rifle forum someone asked about the 6.5x55 (.264) round as a long range sniper round. Which got me thinking. I shoot metallic silhouette competition. The ram silhouette we use is made of steel plate and weighs about 65 lbs. It's not unusual for people to have "ringers", where the bullet makes a good hit but the ram does not go down. Because the ram is steel I would think that the bullet sheds 100% of it's "energy".

I've seen the .264 bullet, going with sufficient velocity knock the ram down, and the .308 bullet - heavier but going slower - fail to knock it down.

No doubt the bullet design, and what it is hitting determines the terminal effect. Real world testing will reveal what that would be, whether on steel or ballistic gel.

But, assuming a bullet is going x fps and shedding 100% of it's energy, is it possible to give a number that tells what force ( such as lbs per square inch) it is hitting an object with? Feel free to use one of the examples in my original post.



It will shed 100% of its energy but that is the wrong way to approach this because this is an inelastic collission. Much of the energy is disspiated in the form of heat in melting the lead, steel plate, noise, etc... so the kinetic energy of the bullet of the bullet does NOT equal the KE of the steel ram after the collision. Do NOT confuse this with the concept of conservation of energy. Conservation of energy says no energy is lost in the total system (The KE of the bullet is never lost, some of it is transformed to heat, noise, etc.. but not all of it is tranferred to moving the ram (KE).

You could calculate the force for an instant of time by multiplying the mass of the bullet by its deceleration but this would be difficult since the bullet is fragmenting and decelerating at different rates. It also wouldn't tell you if it is enough to knock down the ram because the force varies with time and you don't know how that curve looks.

With inelastic collisions, you want to look at momentum. You know what the momentum of the bullet is (mv) and it will be equal to the momentum of the ram after the collision. Since the ram is a swnging target, we can equate the momentum of the bullet to the angular momentum of the ram (moment of inertia x angular veliocity). If you know the moment of inertia of the target, we can then solve for it's angular velocity after the collision. Knowing the angular velocity, we can then go back to dealing with energy. If the ram's rotational KE is greater than the work required to rotate it to the point where it falls down (difference in PE at its normal state and the 12 o'clock position plus energy required ot overcome friction at the hinge/air), then it will fall down.
Link Posted: 12/22/2005 11:57:01 AM EDT
[Last Edit: 12/22/2005 11:58:20 AM EDT by TUMOR]

Originally Posted By DukeSnookems:

Originally Posted By TUMOR:
Kinetic energy = (Mass x Velocity squared)/2



ahhh... ARFCOM

KE= 1/2 x m x v^2, definitely NOT the same as (mv)^2



OK..........you like this better ? KE = ((Mass) X (Velocity Squared))/2



Apology accepted.
Link Posted: 12/22/2005 12:57:42 PM EDT
[Last Edit: 12/22/2005 1:16:18 PM EDT by Mike_Mills]

Originally Posted By Mike_Mills:
It is possible to do such a thing, Pthfndr, but without some real world data it might be hard to believe the answer. For instance:

impulse into target = momentum of bullet

Force on target * duration of impact = mass of bullet * velocity of bullet

Rearranging terms yields:

F = m*v/t

The duration of the collision affects the force. Bullet type (construction, materials,...) and possibly even the actual velocity versus time profile during the collision will affect the results.

P.S. - please, guys, at least get the units correct. It's foot-pounds not foot/pounds, they are different.



Let's do some dumb calculations based on this approach:

Force = mass * velocity / time

mass = (175 gr/7000 gr/lb)/32 ft/sec2 = 0.00078 lbforce sec2/ft (lbmass)

velocity = sqrt(588 ft-lb* 2 /mass) = 1227 ft/sec

momentum = mass * velocity = .00078 * 1227 = 0.96 lb-sec

Assume bullet is 1" long, collision (impulse) time is approximately (dumb first cut),

t = distance/velocity = 1 inch/(1227 ft/sec*12 inch/ft) = 0.000068 sec (68 microseconds)

Force = momentum/time

Force = (0.96 lb-sec)/0.000065 sec) = 14112 lb

Here's a freebie: local pressure on plate's surface = force/area = 14,112 lb/.071 inch2 = 200,000psi

Most steels yield at around 50,000 to 100,000 psi, so now you know why the bullets dent/punch through the target.
Link Posted: 12/22/2005 1:02:38 PM EDT

Originally Posted By Mike_Mills:
Originally Posted By Mike_Mills:

...

Force = momentum/time

Force = (1.00 ft-lb/sec)/0.000065 sec) = 14112 ft-lb



Whoops.
Link Posted: 12/22/2005 1:04:36 PM EDT
I hit that submit button too fast. Edited and updated with a freebie thrown in as an apology for my units error. No one is perfect, especially not me.
Link Posted: 12/22/2005 1:13:24 PM EDT

Originally Posted By TUMOR:

Originally Posted By DukeSnookems:

Originally Posted By TUMOR:
Kinetic energy = (Mass x Velocity squared)/2



ahhh... ARFCOM

KE= 1/2 x m x v^2, definitely NOT the same as (mv)^2



OK..........you like this better ? KE = ((Mass) X (Velocity Squared))/2



Apology accepted.



Link Posted: 12/22/2005 1:22:05 PM EDT

Originally Posted By Mike_Mills:

Originally Posted By Mike_Mills:
It is possible to do such a thing, Pthfndr, but without some real world data it might be hard to believe the answer. For instance:

impulse into target = momentum of bullet

Force on target * duration of impact = mass of bullet * velocity of bullet

Rearranging terms yields:

F = m*v/t

The duration of the collision affects the force. Bullet type (construction, materials,...) and possibly even the actual velocity versus time profile during the collision will affect the results.

P.S. - please, guys, at least get the units correct. It's foot-pounds not foot/pounds, they are different.



Let's do some dumb calculations based on this approach:

Force = mass * velocity / time

mass = (175 gr/7000 gr/lb)/32 ft/sec2 = 0.00078 lbforce sec2/ft (lbmass)

velocity = sqrt(588 ft-lb* 2 /mass) = 1227 ft/sec

momentum = mass * velocity = .00078 * 1227 = 0.96 lb-sec

Assume bullet is 1" long, collision (impulse) time is approximately (dumb first cut),

t = distance/velocity = 1 inch/(1227 ft/sec*12 inch/ft) = 0.000068 sec (68 microseconds)

Force = momentum/time

Force = (0.96 lb-sec)/0.000065 sec) = 14112 lb

Here's a freebie: local pressure on plate's surface = force/area = 14,112 lb/.071 inch2 = 200,000psi

Most steels yield at around 50,000 to 100,000 psi, so now you know why the bullets dent/punch through the target.




OK! That's what I'm talking about. With the understanding that's just a rough example(?)

No way would I want to get with 200K psi. Heck, even if it was just 50K psi I wouldn't want to get hit with it if it was a stream of water from a 1 inch hose, much less a bullet.
Link Posted: 12/22/2005 4:07:49 PM EDT
[Last Edit: 12/22/2005 4:09:36 PM EDT by Andreuha]

Originally Posted By Pthfndr:

Originally Posted By Mike_Mills:

Originally Posted By Mike_Mills:
It is possible to do such a thing, Pthfndr, but without some real world data it might be hard to believe the answer. For instance:

impulse into target = momentum of bullet

Force on target * duration of impact = mass of bullet * velocity of bullet

Rearranging terms yields:

F = m*v/t

The duration of the collision affects the force. Bullet type (construction, materials,...) and possibly even the actual velocity versus time profile during the collision will affect the results.

P.S. - please, guys, at least get the units correct. It's foot-pounds not foot/pounds, they are different.



Let's do some dumb calculations based on this approach:

Force = mass * velocity / time

mass = (175 gr/7000 gr/lb)/32 ft/sec2 = 0.00078 lbforce sec2/ft (lbmass)

velocity = sqrt(588 ft-lb* 2 /mass) = 1227 ft/sec

momentum = mass * velocity = .00078 * 1227 = 0.96 lb-sec

Assume bullet is 1" long, collision (impulse) time is approximately (dumb first cut),

t = distance/velocity = 1 inch/(1227 ft/sec*12 inch/ft) = 0.000068 sec (68 microseconds)

Force = momentum/time

Force = (0.96 lb-sec)/0.000065 sec) = 14112 lb

Here's a freebie: local pressure on plate's surface = force/area = 14,112 lb/.071 inch2 = 200,000psi

Most steels yield at around 50,000 to 100,000 psi, so now you know why the bullets dent/punch through the target.




OK! That's what I'm talking about. With the understanding that's just a rough example(?)

No way would I want to get with 200K psi. Heck, even if it was just 50K psi I wouldn't want to get hit with it if it was a stream of water from a 1 inch hose, much less a bullet.



You'd probably want to get hit even less with water at that pressure. Remember, it's pressure, and a stream of water has much more mass ==> Tons of momentum!

A hose shooting a 1 inch stream at 50kPSI could take out a building. Say, anyone know the operating pressure and nozzle opening diameter of those pattern cutters?
Link Posted: 12/22/2005 4:35:00 PM EDT

Originally Posted By Andreuha:Say, anyone know the operating pressure and nozzle opening diameter of those pattern cutters?


30,000 psi is a normal pressure to "allow the high pressure seal life in the pump to be 700 to 800 hours. At 55,000 psi, which the Paser is capable of, the seal life is 150 to 200 hours."

"As you increase the pressure," Steinhauser said, "you decrease the pump seal life. And maintenance costs go up as you operate the system at higher speeds."

Both companies are working on new nozzles, with 0.0015 to 0.0020 of an inch diameter in the offing. Current diameters are 0.0047 to 0.0067.
Link Posted: 12/22/2005 4:44:12 PM EDT
[Last Edit: 12/22/2005 4:45:53 PM EDT by Andreuha]

Originally Posted By Pthfndr:

Originally Posted By Andreuha:Say, anyone know the operating pressure and nozzle opening diameter of those pattern cutters?


30,000 psi is a normal pressure to "allow the high pressure seal life in the pump to be 700 to 800 hours. At 55,000 psi, which the Paser is capable of, the seal life is 150 to 200 hours."

"As you increase the pressure," Steinhauser said, "you decrease the pump seal life. And maintenance costs go up as you operate the system at higher speeds."

Both companies are working on new nozzles, with 0.0015 to 0.0020 of an inch diameter in the offing. Current diameters are 0.0047 to 0.0067.



Thanks... So I guess even a .224 nozzle could do some damage at 50k PSI. Now just to find a pump capable of delivering that at a (reasonable cost and small enough to mount in the attic)
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