Lug setback? What am I missing?

Forces on the bolt lugs would be the same with the same max chamber pressure. The velocity of the bullet will be a little more. This can be proved if you have a "good" known load with a low ES. Hold it real loose and shoot a few, then hold it super tight or have a big guy shoot it holding tight. You may see 25 fps or more difference.

I have a question...I'm thinking a lead sled and how they are know to break stocks on heavy recoiling rifles. If you held a rifle against a concrete wall and fired it, it would SEEM that more forces are being applied to the entire rifle (in order to break the stock) so more should also be being applied to the bolt lugs?
 
I have a question...I'm thinking a lead sled and how they are know to break stocks on heavy recoiling rifles. If you held a rifle against a concrete wall and fired it, it would SEEM that more forces are being applied to the entire rifle (in order to break the stock) so more should also be being applied to the bolt lugs?
Someone needs to take a crappy old rifle and load up the lightest and heaviest bullets to the same pressure and shoot with the stock against a concrete wall! I love the idea Adam32
 
YEAH! A fun debate :)
So, would that mean recoil is the same at the initial firing of a 110gr bullet at 65k psi and a 230gr bullet at 65k psi, with the DURATION being what causes the increase in felt recoil? Or,
Would it be correct to say the force on the lugs is also the same without regard to the duration ???
Yes I would think that is correct as long as we clarify that the inside case diameter is the same as is everything else about the rifle. When we start getting more dynamic or shift out into recoil mass of objects certainly comes into play. Also, that heavier bullet traveling slower could cause a faster rise in pressure. But at that instant isolating everything...
 
It seems that loss of 25fps would have a decreased force on the bolt lugs.
In these examples the rifle is moving rearward instead of the bullet forward but the same force. Since the bullet is so much lighter, the (and I'm just throwing out examples not calculated numbers) rifle moving rearward at 1ft/sec takes energy that could have added 25ft/sec to the bullet
 
I have a question...I'm thinking a lead sled and how they are know to break stocks on heavy recoiling rifles. If you held a rifle against a concrete wall and fired it, it would SEEM that more forces are being applied to the entire rifle (in order to break the stock) so more should also be being applied to the bolt lugs?
Breaking stocks with the sled is more to do with exceeding the limits of stock strength. Holding the rifle it can "give" a little with recoil in the sled or against a wall not so much.

Here's a test for Adam to try. 🙂 Hit your hand with a hammer just holding it in the air. Now put your hand on a bench or the ground and hit it. Notice the difference in how the same force feels.... 🤣

Sorry, couldn't resist. Just joking with you adam
 
In these examples the rifle is moving rearward instead of the bullet forward but the same force. Since the bullet is so much lighter, the (and I'm just throwing out examples not calculated numbers) rifle moving rearward at 1ft/sec takes energy that could have added 25ft/sec to the bullet
Instead of Force, maybe I should have started with the term Energy. We all know e=mc^2 (capt obvious here). So how does the mass and the acceleration of the bullet not effect the lugs differently??
 
Breaking stocks with the sled is more to do with exceeding the limits of stock strength. Holding the rifle it can "give" a little with recoil in the sled or against a wall not so much.

Here's a test for Adam to try. 🙂 Hit your hand with a hammer just holding it in the air. Now put your hand on a bench or the ground and hit it. Notice the difference in how the same force feels.... 🤣

Sorry, couldn't resist. Just joking with you adam
The "Adam's Hand Test" clearly shows the energy translated to the lugs is different :)
 
Instead of Force, maybe I should have started with the term Energy. We all know e=mc^2 (capt obvious here). So how does the mass and the acceleration of the bullet not effect the lugs differently??
I think that once the case seals. The load transfers to the recoil lug. Unless over pressured
 
Elaborate a bit more...so if recoil is a felt representation of chamber pressure then recoil should be the same across the board with equal chamber pressure? The bullet and case will have zero effect on recoil?

I didn't say recoils was a representation of chamber pressure but the WORK that pressure is doing when combined with a bullet, the same pressure can do varying levels of work depending on the bullet we put in front of it.
 
Let's think of this in reverse, IF recoil was an indication of bolt thrust and therefore chamber pressure it would mean we could take a recoil calculation and it would convert to a chamber pressure. So in practical terms we could take a 300 RUM and load a 150 gr bullet to the same recoil calculation as a 300 gr bullet and also be at the same pressure, well we both know from experience that we would pressure out our action trying to make that happen.
 
Let's think of this in reverse, IF recoil was an indication of bolt thrust and therefore chamber pressure it would mean we could take a recoil calculation and it would convert to a chamber pressure. So in practical terms we could take a 300 RUM and load a 150 gr bullet to the same recoil calculation as a 300 gr bullet and also be at the same pressure, well we both know from experience that we would pressure out our action trying to make that happen.
So BG,
Lets say we load the 150g and 300g to the same pressure. Everything else remains the same. Would the recoil be equal?

Regarding the recoil computation, I think we can make it based on bullet, psi, velocity and rifle weight. We know the energy at the muzzle, thus wouldn't we know the recoil relationship??
 
I have a question...I'm thinking a lead sled and how they are know to break stocks on heavy recoiling rifles. If you held a rifle against a concrete wall and fired it, it would SEEM that more forces are being applied to the entire rifle (in order to break the stock) so more should also be being applied to the bolt lugs?


I have used mine for many years and never had an issue with stock breakage but I don.t add any weight to mine because it already weighs over 20 pounds and extra weight is not needed. It will move slightly but the rifle remains in the same position in the sled as it was, for good repeat ability.

I have added weights for slow motion Video's to get more stable Video, but these were done while testing Muzzle brake designs that reduced recoil.

The only time I have ever seen a stock break, was while shooting a .50 caliber line throwing gun using a 16 oz projectile . We always placed the but plate on a life preserver and held it in place with our knee. some how the but stock slipped off the life preserver and made contact with the deck about the time it fired. The stock didn't survive, :mad: but the sailor did.

J E CUSTOM
 
So BG,
Lets say we load the 150g and 300g to the same pressure. Everything else remains the same. Would the recoil be equal?

Regarding the recoil computation, I think we can make it based on bullet, psi, velocity and rifle weight. We know the energy at the muzzle, thus wouldn't we know the recoil relationship??
If the question is would the recoil be equal in the same gun with two different weight bullets at the same pressure the answer is no because of the bullet weight.
Easy example of that is shoot your 30-06 with 110s & 220s (assuming they are both max pressure) the 220s kick way harder.
Not many of us can accurately test for pressure but using pressure indicators we can prove this to ourselves in our own rifles.

Good discussion has moved from bolt lug strengths to recoil management and back several times.
 
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