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Need help with bolt weight math 22lr
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  gun410999

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Posted: February 28 2017 at 6:39pm | IP Logged Quote gun410999

I posted this in another post with some more questions, but didnt get anyone to answer on that topic. But anyway could someone help me on my 22lr bolt weight? So I read this article on Orions Hammer and put my info in this expression ( bolt mass in pounds = 1.09x10-5 * bullet mass in grains * bullet velocity in fps * (diameter of bolt face / diameter of bullet base)^2) expression with my info 1.09x10-5 *40*1240* (.0625/.226)^2 and i got 0.04134769754 pounds for bolt weight is that right? It dosnt seem right. Thanks
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  backbencher

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Posted: February 28 2017 at 7:29pm | IP Logged Quote backbencher

Dude, use his chart.  Says .4 lbs.

http://www.orions-hammer.com/blowback
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  vintagemx0

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Posted: February 28 2017 at 11:01pm | IP Logged Quote vintagemx0

I'm not sure where you got 0.0625 for the "diameter of the bolt face"? I'm wondering if it was a left-over from an interim calculation. The square root of .0625/pi is equal to a 0.141 radius of a circle. That times two for a diameter is .282. So, maybe you used an area there instead of a diameter? Kinda fat for the base of .22lr, but the closest thing I can imagine for the discrepancy your experiencing.

Looking at the spreadsheet supplied on that link you can see the math and variables that were used. For "bolt face" it is using the "base" (which I assume to be the ID diameter of the casing at the base contacting the bolt face) and for "diameter of bullet base" it is using "caliber". Kinda makes sense...the effective area in one direction and the effective area in the other.

For your value of 40 grains and the spread sheet values of .223 for the caliber and .224 for the base, it would yield :

.0000109 x 1240 x 40 x (.224/.223)^2 = 0.545 pounds

Note: The spreadsheet uses a value of .0000108 and a default weight of 29 grains for the .22lr. These values give you .3919 pounds. The mass of the slug made the bigger difference, but I imagine other variables such as velocity will need to be altered when using specific masses for the slug???

I would recommend taking a close look at the spreadsheet, its formulas, and play around with the variables to really gain any insight to this presentation.

I hope this helps you on your way somehow...

-Ken




















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  junkcollector

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Posted: March 01 2017 at 1:17am | IP Logged Quote junkcollector

This may not be technical enough, but here are some measurements I made of the bolt for a Bingham PPS-50.

Bolt Weight without Lever - 5.3 Oz
Bolt With   Lever     &nb sp;      ;     - 6.0 Oz
Bolt   Diameter          &nb sp;        - .870"
Bolt   Length     &n bsp;    &nbs p;         &nbs p; - 3 5/8"
Bolt Travel Length          &nbs p;   - 1 9/16"

Don't know where the BS with   &   comes from I tried to edit out but it came back, ignore it.
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  gun410999

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Posted: March 01 2017 at 1:58am | IP Logged Quote gun410999

backbencher wrote:
Dude, use his chart.  Says .4 lbs.<br><br>http://www.orions-hammer.com/blowback<br>

<div id="lb" style="display: none; visibility: ; opacity: 0;"><div id="lbCenter" style="display: none;"><div id="lbImage"><a id="lbPrevLink" href="#"></a><a id="lbNextLink" href="#"></a><div id="lbBottomContainer" style="display: none;"><div id="lbBottom"><a id="lbCloseLink" href="#"></a><div id="lbCaption"><div id="lbNumber"><div style="clear: both;">
True but im the type of person that likes to understand things instead of just using someone else's work. And people say that is for closed bolt action mines a open bolt 22lr single shot
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  gun410999

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Posted: March 01 2017 at 2:16am | IP Logged Quote gun410999

vintagemx0 wrote:
I'm not sure where you got 0.0625 for the "diameter of the bolt face"? I'm wondering if it was a left-over from an interim calculation. The square root of .0625/pi is equal to a 0.141 radius of a circle. That times two for a diameter is .282. So, maybe you used an area there instead of a diameter? Kinda fat for the base of .22lr, but the closest thing I can imagine for the discrepancy your experiencing.

Looking at the spreadsheet supplied on that link you can see the math and variables that were used. For "bolt face" it is using the "base" (which I assume to be the ID diameter of the casing at the base contacting the bolt face) and for "diameter of bullet base" it is using "caliber". Kinda makes sense...the effective area in one direction and the effective area in the other.

For your value of 40 grains and the spread sheet values of .223 for the caliber and .224 for the base, it would yield :

.0000109 x 1240 x 40 x (.224/.223)^2 = 0.545 pounds

Note: The spreadsheet uses a value of .0000108 and a default weight of 29 grains for the .22lr. These values give you .3919 pounds. The mass of the slug made the bigger difference, but I imagine other variables such as velocity will need to be altered when using specific masses for the slug???

I would recommend taking a close look at the spreadsheet, its formulas, and play around with the variables to really gain any insight to this presentation.

I hope this helps you on your way somehow...

-Ken
++xxriing ifyour
Thanks that does help, i must have been getting the bolt face diameter mixed up... I put .0625 for bolt face diameter because my build is a open bolt 22lr (single shot) and its gonna have a fixed 1/16 inch firing pin (Drill 1/16 hole in bolt face, put a piece of 1/16 music wire in the hole and have that stick out about a 1/16 from the bolt face to be the firing pin). And i assumed that where it said bolt face diameter it meant the piece of the bolt that's touching the bullet when the gun is fired (firing pin in my case) but i guess not?
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  gun410999

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Posted: March 01 2017 at 2:18am | IP Logged Quote gun410999

junkcollector wrote:
This may not be technical enough, but here are some measurements I made of the bolt for a Bingham PPS-50.

Bolt Weight without Lever - 5.3 Oz
Bolt With   Lever     &nb sp;      ;     - 6.0 Oz
Bolt   Diameter           &nb sp;        - .870"
Bolt   Length     &n bsp;    &nbs p;         &nbs p; - 3 5/8"
Bolt Travel   Length           &nbs p;   - 1 9/16"

Don't know where the BS with   &   comes from I tried to edit out but it came back, ignore it.
Thanks i will take a look at the numbers/specs
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  gun410999

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Posted: March 01 2017 at 9:35pm | IP Logged Quote gun410999

vintagemx0 wrote:
I'm not sure where you got 0.0625 for the "diameter of the bolt face"? I'm wondering if it was a left-over from an interim calculation. The square root of .0625/pi is equal to a 0.141 radius of a circle. That times two for a diameter is .282. So, maybe you used an area there instead of a diameter? Kinda fat for the base of .22lr, but the closest thing I can imagine for the discrepancy your experiencing.

Looking at the spreadsheet supplied on that link you can see the math and variables that were used. For "bolt face" it is using the "base" (which I assume to be the ID diameter of the casing at the base contacting the bolt face) and for "diameter of bullet base" it is using "caliber". Kinda makes sense...the effective area in one direction and the effective area in the other.

For your value of 40 grains and the spread sheet values of .223 for the caliber and .224 for the base, it would yield :

.0000109 x 1240 x 40 x (.224/.223)^2 = 0.545 pounds

Note: The spreadsheet uses a value of .0000108 and a default weight of 29 grains for the .22lr. These values give you .3919 pounds. The mass of the slug made the bigger difference, but I imagine other variables such as velocity will need to be altered when using specific masses for the slug???

I would recommend taking a close look at the spreadsheet, its formulas, and play around with the variables to really gain any insight to this presentation.

I hope this helps you on your way somehow...

-Ken
++xxriing ifyour
And when it says bolt face diameter in the equation does it mean the diameter of the bolt that's touching the round when fired. Cause mine would be 1/16(mines a openbolt 22lr and only the firing pin is touching the bullet when fired. Or is it talking about something else?
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  gun410999

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Posted: March 03 2017 at 7:50pm | IP Logged Quote gun410999

Anyone?
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  PappaSunrise

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Posted: March 03 2017 at 8:27pm | IP Logged Quote PappaSunrise

Use the actual diameter of the bolt face. Remember the firing pin is crushing the case to the point that the bolt face will touch the back of the casing.
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  gun410999

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Posted: March 03 2017 at 11:32pm | IP Logged Quote gun410999

PappaSunrise wrote:
Use the actual diameter of the bolt face. Remember the firing pin is crushing the case to the point that the bolt face will touch the back of the casing.
But then why would the bolt face matter. Because the back of the bullet case can only touch the size of the back of the round, if i used a 3/4" or 5/8"bolt the amount of the casing touching the bolt would be the same. I put my info in and got .0000109 x 1240 x 40 x (.75/.223)^2=6.11 pounds which is mathematically right cause i only changed the bolt face diameter.
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  vintagemx0

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Posted: March 04 2017 at 12:16am | IP Logged Quote vintagemx0

I think the term "bolt face" in this case is a vector referring to the forces at work against the bolt face. The gas pressure pushing against the area on the bolt end of the interior of the casing is what is producing the force which is in turn transferred to the bolt.

Imagine the the hot gases expanding in the chamber. The bullet is being pushed in one direction due to the pressure of the gas pushing on it, and the shell casing is being pushed the other direction by the same pressure. The pressure in PSI is the same on both of ends at any moment, but the force on either end would be dependent on their areas. (Pounds per square inch).

Say the PSI in the chamber at a given moment is 20,000 and let's say the area inside the case at the end adjacent to the bolt is .0028 Sq inches. The force at that given moment imparted to the face of the bolt would be 20,000 x .0028, or 567 pounds of force. It doesn't matter how much of the bolt is contacting the case, it would be 567 pounds of force acting against the bolt regardless.
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  gun410999

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Posted: March 04 2017 at 1:07am | IP Logged Quote gun410999

vintagemx0 wrote:
I think the term "bolt face" in this case is a vector referring to the forces at work against the bolt face. The gas pressure pushing against the area on the bolt end of the interior of the casing is what is producing the force which is in turn transferred to the bolt.

Imagine the the hot gases expanding in the chamber. The bullet is being pushed in one direction due to the pressure of the gas pushing on it, and the shell casing is being pushed the other direction by the same pressure. The pressure in PSI is the same on both of ends at any moment, but the force on either end would be dependent on their areas. (Pounds per square inch).

Say the PSI in the chamber at a given moment is 20,000 and let's say the area inside the case at the end adjacent to the bolt is .0028 Sq inches. The force at that given moment imparted to the face of the bolt would be 20,000 x .0028, or 567 pounds of force. It doesn't matter how much of the bolt is contacting the case, it would be 567 pounds of force acting against the bolt regardless.
Then what am I supposed to enter for the bolt diameter? Should it be .4 pounds or 6 pounds lol(I know it's .4 but when I entered .75 as the bolt diameter I got 6.11 pounds)
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  vintagemx0

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Posted: March 04 2017 at 1:35am | IP Logged Quote vintagemx0

The spreadsheet on Orions Hammer is using the outside diameter of the of the base of the shell casing for "bolt face", so use .224.

You'll get .55 pounds using a 40 grain slug. Your expectation of .4 pounds is based on a 29 grain slug.
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  gun410999

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Posted: March 04 2017 at 2:08am | IP Logged Quote gun410999

vintagemx0 wrote:
The spreadsheet on Orions Hammer is using the outside diameter of the of the base of the shell casing for "bolt face", so use .224.

You'll get .55 pounds using a 40 grain slug. Your expectation of .4 pounds is based on a 29 grain slug.
Ok I'll do the equation for a couple different rounds but I can probally go a little less because of the open bolt momentum.
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  gun410999

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Posted: March 04 2017 at 12:14pm | IP Logged Quote gun410999

.
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  junkcollector

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Posted: March 04 2017 at 2:09pm | IP Logged Quote junkcollector

I'm lost as to the base being .224 " . To me the .224 " is the TUBE diameter of the case, the base to me is the expanded lip at the rear of the shell case which I get as .275 ".
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  gun410999

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Posted: March 04 2017 at 6:33pm | IP Logged Quote gun410999

junkcollector wrote:
I'm lost as to the base being .224 " . To me the .224 " is the TUBE diameter of the case, the base to me is the expanded lip at the rear of the shell case which I get as .275 ".
Yeah i'm a little confused her hopefully someone will chime in again lol.
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  backbencher

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Posted: March 04 2017 at 6:42pm | IP Logged Quote backbencher

In a centerfire case, the bolt thrust area (where the gasses are pressing on the rear of the case) is the area of the rear interior of the case, not the actual casehead diameter.  In the case of the .22" LR, it would be the area of the rear interior of the case at the rim, as the primer burns away and leaves gas pressing against the bolt across the whole rear of the case, including the inside of the rim.

How you calculate rebated cases like the .41" AE & .458" SOCOM I don't know.
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  gun410999

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Posted: March 04 2017 at 9:38pm | IP Logged Quote gun410999

So what bolt weight should i go with? Here are some different rounds i calculated---.0000109 x 1240 x 40 x (.224/.223)^2 = 0.545 pounds- federal 22lr---
.0000109 x 1640 x 32 x (.224/.223)^2 = 0.577 pounds- cci stingers 22lr---
.0000109 x 1280 x 36 x (.224/.223)^2 = 0.507 pounds- winchester 22lr---
.0000109 x 948 x 60 x (.224/.223)^2 = 0.626 pounds- Aguila, .22LR Sniper--- I want to be able to shoot all 3 and if i listen to this i would have to go with a .626 lb or 10 oz bolt. But the guy in this video https://www.youtube.com/watch?v=5a5MyOBQqyY used a 6 oz bolt and it worked with the stingers but he didnt test the aquila rounds(or have them). What should i do?
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