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115gr vs 147gr


scowens

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Not wanting to start a which one is better, but I have a question on the difference between the two. This is an ongoing discussion I am having with a good friend of mine.

Givens:

  • Power Factor is 125,000 for both bullets.
  • same brass, primer, powder.
  • Same bullet; FMJ, RN. Only difference is one is 115gr, and the other is 147gr.
  • Fired from the same pistol
  • Same weather, distance, etc.

Unknown: what is the difference between the two types of bullets?

His stance is that they are vastly different with the 147 less recoil, more knockdown power.

My stance is that if the PF is held the same, other than velocity that each projectile is travelling, they are the same.

I've used the SAMMI Estimated Felt Recoil formula - no significant difference between the two. However when I add the weight difference between 10 115grs vs 10 147grs; there is a difference, but is it significant?

Is there a difference between the two????? And what/why?

Thanks

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Hello: yes there is a difference. The 147 will have more knock down power at the same power factor as the 115's. Inertia is the factor here. Just remember some of your high school physics. The weight difference is 32 grains so you can figure out what the difference will be. If you don't want to do the math load up some of each and shoot a steel popper and see which one knocks it down faster/harder. Hope this helps. Thanks, Eric

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Miranda,

There's a lot more going on when a pistol fires than just energy and momentum; there are two different things being talked about here: recoil impulse and "knock down" power. The first is effected by slide velocity, spring weights, slide stop angles, frictional forces, momentum of moving parts, centers of gravity, etc., the second is heavily dependent on how much energy the bullet can transmit into the target, which is effected bullet weight, composition, the amount of shrapnel, sound, heat, etc.

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Hi kneelingatlas,

the terms and conditions involved are all sorts of splitting hairs.

He asked for a lot more than I think is easily answered on a forum.

I didn't think for a minute that I was writing anything definitive,

rather trying to get a couple of simple answers to point him in the right direction.

Here are some personal fun facts....

I do not seem to be able to tell snappy 115 from pushy 147.

I seem to shoot them about the same.

so I did not speak about that part.

so for fun, lets skip the slide slapping around, the loud booom, the grip factors, and

go to the humorous part

--> just how much energy will a paper target transfer to a flying bullet?

miranda

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--> just how much energy will a paper target transfer to a flying bullet?

I was referring to the energy the bullet transfers to the steel popper :sight:
Because it is not a perfectly elastic collision (between the bullet and the plate) there is no conservation of energy - only conservation of momentum.

A 115 and 147 GR bullet at the same PF have the same momentum (not the same energy). Therefore, whichever bullet leaves the plate (in fragments or as a whole) with less momentum will have transferred more momentum to the plate and thus have more knockdown power (in reference only to steel plates). Not sure that there's an easy model which would dictate which would have more knockdown power but I would guess the difference is minimal compared to things like where the hit is (a hit higher on the popper is more likely to knock over a popper than a hit lower on one) and if the hit was at an angle or straight on.

147s will have less recoil than 115s loaded to the same PF in the same gun IF the same powder is used because more powder is required to get the 115 to the same PF as the 147, and more powder = higher volume of gas leaving barrel at higher speed = more recoil.

Edited by peterthefish
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Make yourself an old school ballistic pendulum and find out which has more juice when it comes to knocking down steel.

;)

An old school ballistic pendulum would not work for that purpose. Bullets bounce off steel and so the momentum of the system is conserved but only the momentum transferred to the steel can knock it down.

A traditional ballistic pendulum captures the bullet allowing the bullet to transfer 100% of its momentum to the pendulum.

Edited by peterthefish
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Generally 115s feel "snappy" while 147s feel more like a push. For newer shooters that push is preferred because the perceived recoil is less. As Jack pointed out, higher level shooters generally don't run 147s because then they run into the situation where they are "waiting for the gun" meaning they are waiting on the slide to cycle before they can get off their next shot. I know that may sound crazy, but it is true once you get to a certain point.

With a lot more powder in a 115 (vs a 147) you generally aren't waiting for the slide. 124s, and more often than not 135s, are the preferred choice because they aren't as snappy and you aren't waiting on the gun as you may be with a 147.

Edited by BAWunder
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As Jack pointed out, higher level shooters generally don't run 147s because then they run into the situation where they are "waiting for the gun" meaning they are waiting on the slide to cycle before they can get off their next shot.

Wait a sec...I know I've read/heard Sevigny, Vogel, and Stoeger all shoot 147's in matches. Am I wrong?

Edited by LeviSS
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No one waits for the slide. The slide is done cycling before a fast split time. Never seen someones trigger finger outrun a pistol.

This. Also, the difference in slide travel time is minimal (<.01) for 115s vs 147s. Even top shooters aren't waiting on the slide to get back for a sight picture.

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"...only the momentum transferred to the steel can knock it down"

I thought that was the original question.

Anyway, it was just a thought.

It's a good thought don't get me wrong. Just saying that a traditional ballistic pendulum captures the bullet - this means it follows a perfectly inelastic collision model, vs bullet vs steel which is imperfectly elastic.

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ok now there are four differences and opinions about the related results.

The differences are cost, trajectory, recoil, and effect on the target.

with most of the discussion being on the last one.

Hi Kneelingatlas,

I was teasing you a bit. I was not sure if you were referring to the linked thread or just this one.

I now think you were talking about the old thread.

once upon a time I took some physics classes and I recall it being a lot of contrived explanations

to show how the math and algebra could help you get to an answer.

so we may want to take a bit of time making sure we agree on a few things

like inertia and mass and energy and momentum and elastic and vectors and acceleration

and velocity and power factor.

I tend to ignore power factor at the target... well that is not true...

I tend to translate it to mass and velocity.

it is easier to explain things.

I would prefer to avoid keeping track of meters per second per second.

here is the easiest way I know to understand what is happening.

If you hit a target like a rock and the bullet flies away with the same speed and the same mass

it had before it hit the rock did the bullet impart any energy to the rock.

or did the rock impart any energy to the bullet?

... well the two did transfer energy...

it is the energy to accelerate to a new flight path.

while there are various limits, in general, a slower and heavier bullet will

transfer more energy compared to a faster and lighter bullet.

this is because the velocity of the bullet did not change...

so it is only the mass of the bullet in the calculation of the acceleration

to the new path....

and the velocity of the bullet dictates how long you have to accelerate to the new path.

miranda

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No one waits for the slide. The slide is done cycling before a fast split time. Never seen someones trigger finger outrun a pistol.

Really? OK, well, google "waiting on the slide 147 9mm". If your splits are in .20s+ you likey won't notice. Below that, well, give it a try and see.

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