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**Test Results** Forward Velocity vs Bullet RPM

@PigButtons brings up a really good point and an excellent analysis of bullet impact of aluminum and water based objects. I need to figure that one out as well in a way that I can print the fragments path and not make a disaster of everything.

ATAI

Would Jello be "fluid" enough to serve your purpose? A pre determined amount in a gallon Ziploc freezer bag lying on its side while setting up should leave you with a sheet of sorts anywhere from a half inch thick or more. Hang the bag in front of your test stand and viola...
 
That certainly makes sense to me! Any ideas or thoughts about the “residue” around the fragment holes? My thought is this is related to hot lead depositing vapor that is trailing it. Clean holes with no residue are either fragments of copper jacket or aluminum. Thoughts? I thought this was another interesting artifact of the experiment.

I certainly can keep posting and testing if people are enjoying it. It’s amazing what you can discover when you try to break things down. Your point about water is well taken...if I can come up with a narrow, fixed width column of water as fragmentation media that would be very interesting. I just want to avoid any hydraulic effect destroying the detection. Maybe gel with card stock molded in every 1/2”. I’ll need to think about that one a bit.

As far as the residue, my thoughts are that the temp of the molten aluminum, is well up into the combustion range upon the impact encounter with the other materials of the bullet. Aluminum is quite combustible, burns very hot, and under certain conditions will ignite as low as 1300* to 1500* F. So as these small fragments of the aluminum plate are traveling through the media stack they are on fire as well. Look at the char around the hole in card #1, and the heat discoloration on the aluminum plate around the holes. Once ignition takes place the aluminum burns at or above 3000* F. Neither lead or copper is likely to combust, steel however can ignite at just 1500*F if you are shooting mild steel targets.

As for test media, what about a piece of thick cow hide like from work boots or belts, or rubber from the sidewall of a tire. Maybe one piece in front of each card in the stack. I'm sure you are a good enough shot that one pair of boots from a thrift store could make 50 1" dia. medalions cut out with a hole saw with no centering bit; enough for 6 tests of 8 cards. You could try some dry and some soaked in water over night.
 
As far as the residue, my thoughts are that the temp of the molten aluminum, is well up into the combustion range upon the impact encounter with the other materials of the bullet. Aluminum is quite combustible, burns very hot, and under certain conditions will ignite as low as 1300* to 1500* F. So as these small fragments of the aluminum plate are traveling through the media stack they are on fire as well. Look at the char around the hole in card #1, and the heat discoloration on the aluminum plate around the holes. Once ignition takes place the aluminum burns at or above 3000* F. Neither lead or copper is likely to combust, steel however can ignite at just 1500*F if you are shooting mild steel targets.

As for test media, what about a piece of thick cow hide like from work boots or belts, or rubber from the sidewall of a tire. Maybe one piece in front of each card in the stack. I'm sure you are a good enough shot that one pair of boots from a thrift store could make 50 1" dia. medalions cut out with a hole saw with no centering bit; enough for 6 tests of 8 cards. You could try some dry and some soaked in water over night.
Interesting, but I’m not so sure about it being aluminum. For one nothing is burned, aluminum burns extremely hot. Second, the lead will melt and vaporize at a much lower temperature than aluminum or copper and this really appears to be a vapor deposit....but it’s pure speculation on my end. Either way I showed that I can make a measurement and that’s good.

Thick wet thick leather is another good material. Making the shot is no problem, I think I’ll gather up a bunch of material and just do some basic test firing. I also think a package of lunch meat (the square packages) backed with some sort of leather to prevent a mess might work as well. Very controllable distance. I’ll keep gathering ideas and test next week when I get back.
 
Would Jello be "fluid" enough to serve your purpose? A pre determined amount in a gallon Ziploc freezer bag lying on its side while setting up should leave you with a sheet of sorts anywhere from a half inch thick or more. Hang the bag in front of your test stand and viola...
This sounds like an interesting idea...my problem is it’s hot as heck here in AZ and it might melt...do you know if the stuff melts?
 
Would Jello be "fluid" enough to serve your purpose? A pre determined amount in a gallon Ziploc freezer bag lying on its side while setting up should leave you with a sheet of sorts anywhere from a half inch thick or more. Hang the bag in front of your test stand and viola...
This sounds like an interesting idea...my problem is it’s hot as heck here in AZ and it might melt...do you know if the stuff melts?
 
Here's a thought why not go the other direction with you test projectiles? Try using more fragile projectiles in those test rifles with reduced loads. Then work up the velocity in increments to see if you develop fragmenting in one rifle before the other. You could also use non metallic medium to initiate fragmentation. Just a thought.
 
Here's a thought why not go the other direction with you test projectiles? Try using more fragile projectiles in those test rifles with reduced loads. Then work up the velocity in increments to see if you develop fragmenting in one rifle before the other. You could also use non metallic medium to initiate fragmentation. Just a thought.
I certainly had this thought as well....just did the initial tests with what I had on hand and was familiar with. Suggestions?
 
Next test "Test 3" coming up:

- 1:8 and 1:16 Twist .223 (300k RPM vs 150k RPM)
- 40g VMax @ ~ 3300fps
- 1" Thick (2" Diameter) piece of dry salami as the initiator media backed with 1/16" thick cardboard
- 3 Cardstock fragmentation detectors spaced 2" apart
- First detection paper at 0.75" from back of cardboard

Thank you for all the ideas for test media! Dry salami wins here in AZ, I can't use anything that will melt or turn to mush in the heat.
 
300 yards =900 feet
900feet / 3300 fps = .272 seconds time of flight to target
300000 RPM= 5000 rps
.272 sec x 5000 rps = 1360 rotations to target
.224" x 3.1416 = .7037" distance around bullet
.7037" x 1360 = bullet spin distance 957"
 
300 yards =900 feet
900feet / 3300 fps = .272 seconds time of flight to target
300000 RPM= 5000 rps
.272 sec x 5000 rps = 1360 rotations to target
.224" x 3.1416 = .7037" distance around bullet
.7037" x 1360 = bullet spin distance 957"
Number of rotations doesn't matter as far as energy being stored/released....rotational energy is stored in the bullet due to rotation once it leaves the barrel, it doesn't require any number of revolutions to release (same as the forward kinetic energy). I assume that is what you are pointing out here.

You calculated the angular distance that the bullet rotates, not the translational distance. Radius of the bullet factors into the rotational kinetic energy equation, not the forward distance per rotation. This is set by the barrel twist. What were you trying to show here? I guess I don’t understand.

Rotations per foot calculation:
Just take the rotations per second r/s (Which you did) and divide by velocity ft/s and you get rotations per foot R/F.

5000/3300=1.515 rotations per foot (or 1 rotation in 8”)
 
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Since we're all trying our hand at high school math: (I'm going to round off some #'s so it is more easily done.)

~ 3000 fps / 2" ( distance between detectors ) or ~ 0.2 feet. = 15000 (15K detector transits per second ).

5000 rps / 15000 transits = 0.33333 rotations between two detector sheets.

Now why is this important? I have no idea, but it's just fun to think about.

Maybe I should have done degrees of rotation, or better yet radians.
 
Since we're all trying our hand at high school math: (I'm going to round off some #'s so it is more easily done.)

~ 3000 fps / 2" ( distance between detectors ) or ~ 0.2 feet. = 15000 (15K detector transits per second ).

5000 rps / 15000 transits = 0.33333 rotations between two detector sheets.

Now why is this important? I have no idea, but it's just fun to think about.

Maybe I should have done degrees of rotation, or better yet radians.
I think you just created a new unit...Detector Transits / Second.

A much, much easier way is to take the distance in between detectors and divide by the barrel twist. 2/8 = 0.25 rotations per detector

You must have used a 1:6 twist barrel in your test. ;)

A Varmint bullet doesn’t drill holes...it explodes! ;)
 
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Fun fact:

It doesn’t matter how fast or slow a bullet travels it will always rotate at your barrel twist rate.

1:8 twist means 1 revolution for every 8” of forward travel

simple but easy to overthink.
 
Number of rotations doesn't matter as far as energy being stored/released....rotational energy is stored in the bullet due to rotation once it leaves the barrel, it doesn't require any number of revolutions to release (same as the forward kinetic energy). I assume that is what you are pointing out here.

BTW the circumference of a bullet is 2xPIxradius
You calculated 2x the angular distance that the bullet rotates, not the translational distance. Radius of the bullet factors into the rotational kinetic energy equation, not the forward distance per rotation. This is set by the barrel twist. What were you trying to show here? I guess I don’t understand.

Rotations per foot calculation:
Just take the rotations per second r/s (Which you did) and divide by velocity ft/s and you get rotations per foot R/F.

5000/3300=1.515 rotations per foot (or 1 rotation in 8”)

Wasn't really trying to show or prove anything.. Was just giving some (probably useless) information..
With regards to your formula about the "C"ircumference ( or distance around the bullet)
C=2xPixR or C=Pi x D Same result...
I always thought a 223 was .224" in diameter...
All is was showing was the total rotational distance the bullet traveled during its flight.. Again useless information..
Not wanting to start a battle about your calculations or work on this.. so, I wish you luck and success in your efforts..
 

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