Loads For Knocking Stuff Down.

[Dowter]

I looked for this info elsewhere and couldn't find it.

What is the general philosophy behind knocking stuff down (pins, poppers,etc.) in regards to bullet weight.

If I want to knock stuff down and with the same power factor, should I go for a heavier or lower weight bullet? In theory it seems like the energy that something is hit with should be the same if the pf is the same and that something should fall down equally as well. (Of course reality can be cruel when it comes to people's theories.)

I had always heard people talking about these real heavy and powerful loads for knocking down pins (I'm assuming bowling pins). I wondered why this was done since I never had any problem with knocking 'em down with my 180gr 170pf loads.

I want about a 130-140 pf load for knocking down poppers.(shoot-offs,etc.). Should I stay at 180gr or go lighter?

Thanks for everyone's help in advance.

[Patrick Sweeney]

For pins, I can speak mostly about Second Chance. Maybe we were manyly men, and a self-selected group, but power came with bulltes int he middle to upper range of each caliber. Sometimes beyond, as with Jerry Miculek and his 205 to 230 gr .357 magnum bullets. SC was timed to the last pin down, or the last pin strikign the backstop behind the tables. If you broomed a pin off straight back you could save a tenth or two by hitting the wall.

Only near the end did we start seeing guys with Supers using 115 and 125's loded to the gills taking pins off with 180PF. But they were centerpunching pins with Open guns.

If there was no power factor, we were shooting falling steel, and timed by sound, I think I'd be running somehting in the 140 range. For pins, 195 or more.

[Flexmoney]

It is the transfer of energy that knocks stuff down.

The heavier bullets tend to transfer the energy into the target better...less likely to glance off.

Bullet shape plays a part as well. Bullets that "stick" to the pins...or "dwell" on the steel will tend to transfer more of their energy into the target.

[Dowter]

Bullet shape plays a part as well. Bullets that "stick" to the pins...or "dwell" on the steel will tend to transfer more of their energy into the target.

So hollowpoints are probably better than FMJs?

[Flexmoney]

So hollowpoints are probably better than FMJs?

That's my thought.

Sierra designs bullets with some of this in mind for silhouette shooters.

[TBF]

Just my .02, Heavy is better for pins. Also SWC or flatnose design better than roundnose for

pins. Seems like 800 fps or above lessens ricochet problems. Most any load 170+ will take off a pin with a

good hit, but more is better if the hit is not perfect.

Travis F.

[Norm Lee]

Hey Dowter:

When it comes to knocking stuff down, it's the transfer of momentum that does the job. Since PF is an analog of momentum, it really won't much matter whether you choose a light or a heavy bullet. Because you will have a higher energy proposition when shooting lighter bullets at the same PF, you may notice a different recoil signature and this may be what might motivate you to a heavier bullet. For pins, which must be persuaded to leave the table entirely, most agree that a minimum PF of 190 or 195 is required. We encourage our pinshooters to keep the velocity above 850 fps to mitigate bounce back. Many have found that velocities above about 1450 result in too much penetration, too soon which interferes in lots of cases with the momentum dump desired. Given these constraints, we see more heavy bullets than light. Hollow points, SWC and WC shapes tend to be more efficient in the transfer than fmj.

For steel, calibrated to fall at minor PF, I like to use lighter bullets, generally. Usually I just swap a 155 for the 185 my usual IPSC .40 load and all is well. In the .45, I like a light load using a 200 gr.swc.

Cheers,

Norm

[Carlos]

I agree w/ TBF's comments and there just happens to be a listed book load for .45 ACP that fits the bill and stays within SAAMI specs. Pay $15 or so for Richard LEE's reloading handbook and look up the data for either Hodgdon HP-38 or Winchester 231 (use only the powder listed) and lead bullets of either 255 or 260 grains. I used the Bull-X lead 255 SWC meant for revolver use; it has a pronounced crimp groove. Charge as directed and crimp at the crimp groove; I got exactly what the book predicted on the chrono: 800 FPS out of an STI Edge (204 PF). Ideally. velocity would be a bit higher to prevent those pesky "low-velocity rollbacks." And yes, Virginia, 255s ricocheting off pins DO hurt when they hit you!

TBF wrote: "Just my .02, Heavy is better for pins. Also SWC or flatnose design better than roundnose for

pins. Seems like 800 fps or above lessens ricochet problems. Most any load 170+ will take off a pin with a

good hit, but more is better if the hit is not perfect."

[shred]

Agreed on heavy bullets for pins, especially in smaller calibers. In 38 Super, I found 147gr Golden Sabers much more effective at the same PF (180+) than 115 JHPs-- probably because they stay in the pin better for that all-important energy transfer, although a good 180gr .40 JHP bullet seemed just as effective as a 200 or 220gr at the same PF.

For knocking down stuff like steel plates, I go with lighter bullet just so they get there faster. At 25 yards, there's a somewhat significant flight-time delay between, say a 147 at 750 fps and 90gr at 1200 fps.. 0.04 sec doesn't sound like a lot, but it can add up in a hurry.

(note that pins are closer to an inelastic collision and steel is closer to an elastic collision-- High-school physics says you should need less energy to knock down some steel the same weight as a pin)

[tightloop]

If you are shooting steel, you need a pf that will take it down with a hit in the lower part of the A zone, probably 135/140.

If you are shooting pins, the much bigger pf most guys shoot is due to the fact that at a real Pins match you must not only knock them down, you must drive them off the table, which is much harder than just knocking them over. You will see pf at 200 or a little more at matches like this, not everyone, but some.

[Dowter]

OK I think that I got it

Same PF = Same Energy/Momentum

BUT

Larger/Slower transfer the energy to a falling plate better(less ricochet) meaning that you are more likely to knock down the popper

BUT

Smaller/Faster gets to the plate faster (any maybe cycles the gun faster?)

AND ALSO

SWCs or Hollow Points transfer energy better than jacketed bullets

So...

for poppers - larger/slower if you're concerned about knocking them down

for poppers - smaller/faster if you're not worried about knocking them down and are looking for speed.

So I guess for Steel Challenge Steel

static steel - smaller/faster (don't care about energy transfer)

How does that sound

[tightloop]

Pretty close.

[Norm Lee]

Hey Dowter:

I agree . Pretty close except for :

"Same PF = Same Energy/Momentum"

If loads have the same PF (or momentum), the one with the lighter bullet will have more kinetic energy. This is because the velocity term in the energy calc is squared. Momentum is simply M*V.

cheers,

Norm

pedantics r us

[Dowter]

Ok, I just looked up the formula.

Kinetic Energy = 1/2 Mass * Velocity^2

180gr at 1000fps (180pf)

KE=90*1000^2

KE=90*1000000

KE=90million

135gr at 1333fps (180pf)

KE=67.5*1333^2

KE=67.5*1777777.77

KE=120million

Is that right?

[Dowter]

Ok, I know I'm missing some bit of physics knowledge that would make this an easy question but I gotta ask.

If the same PF(180) for two different weight bullets produce different ammounts of kinetic energy wouldn't the bullet that is lighter cause the gun to recoil more since it is creating more kinetic energy?

[Norm Lee]

Hey Dowter:

Good example. Units are a little askew but that doesn't really matter. And you've asked a really good question. Might get some lively discussion here.

My take is this. If the PF is the same, so is the recoil, since conservation of momentum is the principle used to figure the recoil. The wild card is that the amount of powder used, converted to gas, is generally figured to be travelling at a pretty fair rate of knots.(faster than the projectile!) You can look up the formula in your reloading manual. Thus, the momentum product for a small difference in powder charge can actually be more significant than you might guess. It's probably the reason it's pretty popular practice to load small weights of fast powder behind heavy bullets to achieve a given power factor. Lots of us are only looking for a little edge http://www.brianenos.com/forums/html/emoti...ons/biggrin.gif

This practice is often frowned upon by conservative reloaders as pressures can spike severely under these conditions.

The other thing, as I noted in an earlier post is that the recoil signature will be different because of the energy difference.

But if you have two loads where the PF is the same and the powder charge is too. then you won't be able to tell the difference between them even if the bullets are of different weights. At least, that's what I think.

Cheers,

Norm

Short answer to your question is yes (but qualified by dissertation above)

[kaiserb]

Slow Heavy....

Big Mass transfers big energy.

[TBF]

I am not an engineer, and I will not offer a math lesson. But big and heavy is better for pins.

Example: my 357 w/125gr bullet at 1650fps has more energy than my 45 w/230gr. bullet.

But the 45 throws the pins better and further . Travis F.

[Dowter]

I submitted this question to some "ask a physics expert" websites. I'll post whatever results I get back.

----------------------

I belong to a group of hobby pistol shooters and we shoot at steel targets and try to knock them over. There is much debate over what is the most efficient bullet design to knock over a steel plate.

There are two competing formulas used. Both want to have as little recoil as possible while transfering as much energy to a steel plate to knock it over.

The first is momentum which is weight * velocity. The other is kinetic energy which is 1/2 weight * velocity^2. Which formula should be used to properly determine how well a plate will be knocked over? Also which formula is best to determine how much recoil will be fealt from the pistol of the shooter?

Another argument that has come up is about bullet weight. It is argued that at the same momemtum or kinetic energy a heavy/slow bullet will transfer energy to a steel target better than a light/fast bullet because the more mass - the better the transfer of momentum or kinetic energy.

Thank you for your help.

[kaiserb]

Here is a good link

http://www.phys.virginia.edu/classes/252/e...rgy_p_reln.html

Just rember the weight is different than mass.

[Norm Lee]

I guess I wasn't as clear as I should have been.

Momentum is conservative. That means the M*V of the projectile transforms to the (M+m)*v of the thing which is hit. For example a 230 gr. projectile at 900 fps strikes a bowling pin (and sticks in). The new combo weighs 3.5 pounds plus 230 grains and will move at about 8.37 fps.(neglecting friction effects)

Momentum transfers to plates too. A simplifying assumption has the bullet simply stop, then the plate moves. The collision is more elastic in the real world and the math is more tricky but it's not really necessary to assess what happens nor to determine knockdown effectiveness.

So I would maintain that if the PF is the same, so is the performance, on things you wish to knock down.

There are other reasons to favor light or heavy bullets which have to do with how effectively they couple with the target (for efficient momentum transfer) and the recoil effects.

Given a choice, I use heavy bullets for pins and light ones for steel.

Cheers,

Norm

Energy doesn't transfer conservatively. Some will invariably dissipate as heat, light, sound, etc.

[Dowter]

Here is a response that I got from a scientist

---------------------------------------------------------

Vincent,

The problems you pose are interesting. Yes, you do want to

minimize the recoil of the gun. Less recoil means more energy

for the bullet. It doesn’t matter which equation you use (MV

or 1/2MV^2) to evaluate what happens to the plate. The

momemtum equation is easier to work with but the energy

equation provides the same result. Both assume ideal

inelastic, frictionless environments, which do not exist in

the real world. So, the answer to your problem lies in

looking at the friction and collision characteristics of both

bullets. The smaller bullet will have a higher speed -

assuming the same gun-powder charge for both bullets.

Aerodynamic drag increases exponentially with an increase in

speed. Thus the faster bullet loses more kinetic energy

(speed) to heat due to air friction. This assumes both have

the same frontal cross-sectional area. If the more massive

bullet has a greater cross sectional area -the increased

difference in area is not as great a determining factor to

drag as speed. Thus, the faster bullet will lose more energy

due to drag. Secondly, what happens to each bullet upon

collision is important in determining a knock down. Bullets

that do not fragment will transfer more energy to the target.

Thus, if the smaller bullet fragments the larger bullet gets

the advantage. If the larger bullet fragments the smaller

bullet has the advantage. Secondly, I imagine there is some

deformation of the metal plate. The smaller bullet because of

its smaller cross sectional area will produce a greater force

on a smaller area of the plate. This allows for greater

deformation of the plate and thus more energy to be lost in

the collosion by the smaller bullet. All of this assumes that

the gun powder charge for both bullets is equal. The more

massive bullet has an advantage, not becasue of it’s mass

directly, but the other isssues discussed above. Hope this helps.

Gregg Zulauf

[TBF]

TIME,

Not only my favorite Pink Floyd song , but a variable missing in most of these calculations.

Recoil:

If you assume that perception of recoil is the same for any instance using only MV and

ejecta ( bullet and powder ) as variables, try this:

300 win. mag same power factor load, very light bullet vs. very heavy bullet, hold tight.

Now do the same experiment, this time holding about an inch off your shoulder.

Due to the TIME variable you will notice a distinct difference in PERCIEVED recoil,

the total recoil will have remained the same.

Travis F.