testing bulletproof glass - DiscussWorldIssues - Socio-Economic Religion and Political Uncensored Debate

**nTDsD0aU** · 10-29-2009, 08:54 AM

I'm thinking BS. She could not hold on tight enough to stop the force of the bullet.

**9V42h1eT** · 10-29-2009, 09:48 AM

I'm thinking BS. She could not hold on tight enough to stop the force of the bullet.

The force on of the bullet is equal to or less than the force of recoil.

**indartwm** · 10-29-2009, 09:53 AM

I'm thinking BS. She could not hold on tight enough to stop the force of the bullet.

It takes less power to stop the bullet than in receiving the recoil at firing it.

Very brave or very stupid woman anyway, especially regarding taking on several shots on the same panel of glass. That's idiocy.

**OWV9LSxH** · 10-29-2009, 09:57 AM

That is freaking hilarious [rofl]

**RobsShow** · 10-29-2009, 10:07 AM

It takes less power to stop the bullet than in receiving the recoil at firing it.

Very brave or very stupid woman anyway, especially regarding taking on several shots on the same panel of glass. That's idiocy.

The force on of the bullet is equal to or less than the force of recoil.

You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact.

**indartwm** · 10-29-2009, 10:18 AM

You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact.

The impact was taken on dampened by the panel itself and the woman's arms. Think about if she'd kept the panel of glass eg. against her forehead directly behind the hit. Ouch...

**UriyVlasov** · 10-29-2009, 10:21 AM

You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact.

Assume no frictional/parasitic loss...

An impulse was provided to the bullet from the expanding gas. The force that projected the bullet is equal and opposite to the force provided to the hand gun.

Since we ignore loss, the momentum of the bullet is completely conserved until it hits the glass, at this point the bullet contributes an impulse to the glass equal to its momentum over the period of time it went from top speed to stopped.

You can assume that the time from gun powder combustion to the bullet leaving the barrel is approximately equal to the time that the bullet hit the glass to when it stopped moving forward.

Same force, same impulse... conservation of momentum.

**ReggieRed** · 10-29-2009, 10:26 AM

Assume no frictional/parasitic loss...

An impulse was provided to the bullet from the expanding gas. The force that projected the bullet is equal and opposite to the force provided to the hand gun.

Since we ignore loss, the momentum of the bullet is completely conserved until it hits the glass, at this point the bullet contributes an impulse to the glass equal to its momentum over the period of time it went from top speed to stopped.

You can assume that the time from gun powder combustion to the bullet leaving the barrel is approximately equal to the time that the bullet hit the glass to when it stopped moving forward.

Same force, same impulse... conservation of momentum.

Pretty much!

**fiettariaps** · 10-29-2009, 10:48 AM

The olden people had balls.

**freeringsf** · 10-29-2009, 10:49 AM

I still like the ones that test the chain mail for the shark suits the best.

**seodiary** · 10-29-2009, 11:10 AM

The olden people had balls.

Even the women...

Such times.

**SpecialOFFER** · 10-29-2009, 11:19 AM

Even the women...

Such times.

In Soviet Russia (esp East Germany), the women f00k YOU! [thumbup]

**anatmob** · 10-29-2009, 11:34 AM

You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact.

How could you expect the bullet to have more energy impacting than what was imparted at the time of firing. The bullets do the damage they do because they are small. the recoil doesnt give the same sense of the momentum imparted on the bullet because you can hold a gun firmly and dissipate the recoil easier over a larger surface area.

**SpecialOFFER** · 10-29-2009, 11:47 AM

So... If I have a .30 caliber rifle shooting a 220gr bullet @ 3300fps, how much energy would the bullet have in foot-pounds at the muzzle, and how many foot-pounds of recoil would have to be absorbed by the shooter?

EDIT: Oh, out of a 26", 1-in-11.4" twist moly-coated tube... in case you needed that. Bullet's moly-coated with a ballistic coefficient of .629 and sectional density of .331... I've also got them w/ .631 and .323.

**mirvokrug** · 10-29-2009, 12:51 PM

Oh, btw - that looks like the Topperweins in the video, so you're likely looking at ~1920-1930...

**Elisabetxxx** · 10-29-2009, 01:45 PM

*Ahem...*

Bullet resistant glass.

10-29-2009, 08:54 AM	#1
nTDsD0aU Join Date Oct 2005 Posts 528 Senior Member	testing bulletproof glass I'm thinking BS. She could not hold on tight enough to stop the force of the bullet. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 09:48 AM	#2
9V42h1eT Join Date Oct 2005 Posts 430 Senior Member	I'm thinking BS. She could not hold on tight enough to stop the force of the bullet. The force on of the bullet is equal to or less than the force of recoil. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 09:53 AM	#3
indartwm Join Date Oct 2005 Posts 429 Senior Member	I'm thinking BS. She could not hold on tight enough to stop the force of the bullet. It takes less power to stop the bullet than in receiving the recoil at firing it. Very brave or very stupid woman anyway, especially regarding taking on several shots on the same panel of glass. That's idiocy. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 09:57 AM	#4
OWV9LSxH Join Date Oct 2005 Posts 519 Senior Member	That is freaking hilarious [rofl] Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 10:07 AM	#5
RobsShow Join Date Oct 2005 Posts 466 Senior Member	It takes less power to stop the bullet than in receiving the recoil at firing it. Very brave or very stupid woman anyway, especially regarding taking on several shots on the same panel of glass. That's idiocy. The force on of the bullet is equal to or less than the force of recoil. You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact. Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 10:18 AM	#6
indartwm Join Date Oct 2005 Posts 429 Senior Member	You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact. The impact was taken on dampened by the panel itself and the woman's arms. Think about if she'd kept the panel of glass eg. against her forehead directly behind the hit. Ouch... Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 10:21 AM	#7
UriyVlasov Join Date Oct 2005 Posts 511 Senior Member	You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact. Assume no frictional/parasitic loss... An impulse was provided to the bullet from the expanding gas. The force that projected the bullet is equal and opposite to the force provided to the hand gun. Since we ignore loss, the momentum of the bullet is completely conserved until it hits the glass, at this point the bullet contributes an impulse to the glass equal to its momentum over the period of time it went from top speed to stopped. You can assume that the time from gun powder combustion to the bullet leaving the barrel is approximately equal to the time that the bullet hit the glass to when it stopped moving forward. Same force, same impulse... conservation of momentum. Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 10:26 AM	#8
ReggieRed Join Date Oct 2005 Posts 491 Senior Member	Assume no frictional/parasitic loss... An impulse was provided to the bullet from the expanding gas. The force that projected the bullet is equal and opposite to the force provided to the hand gun. Since we ignore loss, the momentum of the bullet is completely conserved until it hits the glass, at this point the bullet contributes an impulse to the glass equal to its momentum over the period of time it went from top speed to stopped. You can assume that the time from gun powder combustion to the bullet leaving the barrel is approximately equal to the time that the bullet hit the glass to when it stopped moving forward. Same force, same impulse... conservation of momentum. Pretty much! Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 10:48 AM	#9
fiettariaps Join Date Oct 2005 Posts 419 Senior Member	The olden people had balls. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 10:49 AM	#10
freeringsf Join Date Oct 2005 Posts 501 Senior Member	I still like the ones that test the chain mail for the shark suits the best. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 11:10 AM	#11
seodiary Join Date Oct 2005 Posts 488 Senior Member	The olden people had balls. Even the women... Such times. Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 11:19 AM	#12
SpecialOFFER Join Date Oct 2005 Posts 613 Senior Member	Even the women... Such times. In Soviet Russia (esp East Germany), the women f00k YOU! [thumbup] Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 11:34 AM	#13
anatmob Join Date Oct 2005 Posts 598 Senior Member	You guys are right, but about the wrong thing. We are not talking about the bullet leaving the gun, we are talking about impact. How could you expect the bullet to have more energy impacting than what was imparted at the time of firing. The bullets do the damage they do because they are small. the recoil doesnt give the same sense of the momentum imparted on the bullet because you can hold a gun firmly and dissipate the recoil easier over a larger surface area. Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 11:47 AM	#14
SpecialOFFER Join Date Oct 2005 Posts 613 Senior Member	So... If I have a .30 caliber rifle shooting a 220gr bullet @ 3300fps, how much energy would the bullet have in foot-pounds at the muzzle, and how many foot-pounds of recoil would have to be absorbed by the shooter? EDIT: Oh, out of a 26", 1-in-11.4" twist moly-coated tube... in case you needed that. Bullet's moly-coated with a ballistic coefficient of .629 and sectional density of .331... I've also got them w/ .631 and .323. Share Share this post on Digg Del.icio.us Technorati Twitter
	Quote

10-29-2009, 12:51 PM	#15
mirvokrug Join Date Oct 2005 Posts 436 Senior Member	Oh, btw - that looks like the Topperweins in the video, so you're likely looking at ~1920-1930... Share Share this post on Digg Del.icio.us Technorati Twitter
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10-29-2009, 01:45 PM	#16
Elisabetxxx Join Date Oct 2005 Posts 541 Senior Member	Ahem... Bullet *resistant* glass. Share Share this post on Digg Del.icio.us Technorati Twitter
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