Stupid question. What is "soft setting" a target?

[Zeneffect]

https://www.sacvalley.org/clubs/semi-auto-military-rifle-silhouette/

I want to do this sort of competition. Seems fun and something I could be well suited for. I found out today that at the 550yd(ish) mark I'm not knocking down the rams consistently with a 130gr norma GT doing 2465 at the muzzle. I didn't set them up so I'm not sure how they are placed but the rules (linked) say it's acceptable to soft set them. I don't know what this means exactly, can anyone elaborate? Is it like sort of hanging them at the edge of their platform so it's easier to knock down?

[Old Bob]

Long time black powder cartridge silhouette shooter here. Yup, soft set means the rams are set as far back on their stands as possible without them falling off. It's a delicate balance. It's hard to do it on a windy day...

[LR1955]

https://www.sacvalley.org/clubs/semi...le-silhouette/

I want to do this sort of competition. Seems fun and something I could be well suited for. I found out today that at the 550yd(ish) mark I'm not knocking down the rams consistently with a 130gr norma GT doing 2465 at the muzzle. I didn't set them up so I'm not sure how they are placed but the rules (linked) say it's acceptable to soft set them. I don't know what this means exactly, can anyone elaborate? Is it like sort of hanging them at the edge of their platform so it's easier to knock down?

ZE:

The problem is your load is either too weak or your accuracy is not sufficient for a solid hit. Looking at your velocity I bet it is insufficient KE. Try some 140's?

I wouldn't want someone rigging the targets for me or any competitor because it defeats the intent behind metallic silhouette shooting which is using a cartridge powerful enough to put the steel down given a decent hit.

LR-55

[Zeneffect]

It wasn't competition, just some fun shooting. Targets were preset when I got there so no chance to see how they were set. In competition I would most definitely set my own. I think just not enough power to knock over the ram as I was very clearly hitting center mass and could see the marks very clearly (or maybe aiming for center was the problem?)

Ballistics calculator is showing it should be about 830fpe at target, and I cant do the math to see what that does against a 20kg+ steel ram (is there a good
Formula or general guideline?)

*edit* oh what google can do.

"The theoretical minimum for Ram knockdown is a Terminal Momentum(TM) of 0.7; the closer to 1.0 the better. TM is calculated by multiplying the remaining velocity at 200m (FPS) x bullet weight (grains) divided by 225200 (constant). Ex. 140 Grain bullet w/ a muzzle velocity of 1758 FPS and a 200m velocity of 1459 FPS = 1459 x 140 / 225200 = TM of 0.91." now i need to figure out this math lol. im a stupid person.

[LR1955]

It wasn't competition, just some fun shooting. Targets were preset when I got there so no chance to see how they were set. In competition I would most definitely set my own. I think just not enough power to knock over the ram as I was very clearly hitting center mass and could see the marks very clearly (or maybe aiming for center was the problem?)

Ballistics calculator is showing it should be about 830fpe at target, and I cant do the math to see what that does against a 20kg+ steel ram (is there a good
Formula or general guideline?)

*edit* oh what google can do.

"The theoretical minimum for Ram knockdown is a Terminal Momentum(TM) of 0.7; the closer to 1.0 the better. TM is calculated by multiplying the remaining velocity at 200m (FPS) x bullet weight (grains) divided by 225200 (constant). Ex. 140 Grain bullet w/ a muzzle velocity of 1758 FPS and a 200m velocity of 1459 FPS = 1459 x 140 / 225200 = TM of 0.91." now i need to figure out this math lol. im a stupid person.

I get .75 flb/s if I swag a very liberal 1300 fps velocity at 550 yards. I would think a guy would want the result based on the true distance of 500 M and not 200 M.

I used a mass calculator and not that formula you found. I am pretty sure it will correspond to that really strange formula you found.

When you figure out your velocity at 200M, how about posting your result and lets see if we are close?

LR55

[Zeneffect]

If I plug in values for 550yds, I'm in the .97 range so I'm either doing it wrong, or I'm doing it wrong. I can share my Ballistics chart if that helps?

This is the 130GR Norma with 29.7grLVR that I was shooting.

Distance Speed f/s Energy, ft-lbf Elevation, MOA
100.0 2312.3 1543 D4.8
150.0 2238.2 1446 D4.2
200.0 2167.0 1355 D3.0
250.0 2095.6 1268 D1.6
300.0 2025.6 1184 D0.1
350.0 1956.7 1105 U1.6
400.0 1890.4 1032 U3.4
450.0 1823.8 960 U5.3
500.0 1758.2 892 U7.3
550.0 1695.0 829 U9.5
600.0 1631.4 768 U11.7
650.0 1568.9 710 U14.1
700.0 1507.4 656 U16.7
750.0 1448.3 606 U19.3
800.0 1389.2 557 U22.2
850.0 1331.3 512 U25.2
900.0 1276.2 470 U28.3
950.0 1221.4 431 U31.7
1000.0 1168.9 394 U35.3

[grayfox]

Using my Point Blank, 2465 f/sec MV with 0.548 BC bullet (130 norma gt, 6.5mm), 130 gr...
and my calculator...

at 550 yd, vel= 1713 fps, ke= 847ft-lb.

Using your TM equation with 550 yd data, the knockdown comes to 0.9888, is what I get. I'm calculating this at 500 ft elev.
Using the 200 m figure (or ~220 yds since I can't do m with my app), it would be much higher b/c the Vel would be higher.

[Zeneffect]

doing some further up on this and it seems general consensus is 1000fpe is considered appropriate for the rams. norma 130GT just wont get me there with my gun I suppose. the Berger 130gr ar hybrids get me a bit closer though, however i cant find any more and have maybe 15 left total. guess its worth a try.

[grayfox]

[LR1955]

If I plug in values for 550yds, I'm in the .97 range so I'm either doing it wrong, or I'm doing it wrong. I can share my Ballistics chart if that helps?

This is the 130GR Norma with 29.7grLVR that I was shooting.

Distance Speed f/s Energy, ft-lbf Elevation, MOA
100.0 2312.3 1543 D4.8
150.0 2238.2 1446 D4.2
200.0 2167.0 1355 D3.0
250.0 2095.6 1268 D1.6
300.0 2025.6 1184 D0.1
350.0 1956.7 1105 U1.6
400.0 1890.4 1032 U3.4
450.0 1823.8 960 U5.3
500.0 1758.2 892 U7.3
550.0 1695.0 829 U9.5
600.0 1631.4 768 U11.7
650.0 1568.9 710 U14.1
700.0 1507.4 656 U16.7
750.0 1448.3 606 U19.3
800.0 1389.2 557 U22.2
850.0 1331.3 512 U25.2
900.0 1276.2 470 U28.3
950.0 1221.4 431 U31.7
1000.0 1168.9 394 U35.3

Ze:

Well, if you are getting .97 and it isn't going over, your calculation is wrong. Just not sure how wrong. That equation is bizarre because they seem to be taking a value at 200 then using a constant to estimate a figure at 500 when all they need to do is figure it out at 500 M (550 yards).

I bet the calculator I used is more correct but I need to learn a lot more about how mass is used for this purpose. Too bad someone can't just come up with a simple thing like a Power Factor where you would use velocity at the target, sectional density, and bullet weight. I bet bullet construction has a lot to do with it too. And a bullet that retains its mass when it hits will exert more force than a bullet that disintegrates.

A lot of stuff to consider.

LR-55

[Zeneffect]

I ran my calculation a few times, and what I think is the problem is the equation is garbage and google is full of lies.

Perhaps step 1 is to just do a simple retest, but set the targets myself then evaluate my failures once variables are removed. This should confirm I can use what I have, or need to spend money.

[grayfox]

yeah you guys are probably right, garbage equation. Should have suspected that when it talked about meters but used some SAE values, f/sec, grs, etc.

It's a physics problem of ke or momentum, but the equation doesn't hit me yet.
Would need to know the mass of the ram, and the point of impact, need the thickness or depth of the base of the ram also (ie, front to back measurement at the base), ht may might be helpful.
Takes less ke to "knock it over" when hitting the top than when hitting close to the bottom.

[LR1955]

yeah you guys are probably right, garbage equation. Should have suspected that when it talked about meters but used some SAE values, f/sec, grs, etc.

It's a physics problem of ke or momentum, but the equation doesn't hit me yet.
Would need to know the mass of the ram, and the point of impact, need the thickness or depth of the base of the ram also (ie, front to back measurement at the base), ht may might be helpful.
Takes less ke to "knock it over" when hitting the top than when hitting close to the bottom.

GF:

I went to a physics site and used a calculator they had that gave different measurements of mass. It is legit.

I am not sure if flbs/s is the right answer though.

They shoot for legs and high shots when using the light stuff I think. Given ZE's competition allows every target to be shot from the prone using bags, rests, et. al, shot placement becomes something more than an excuse for failing to bring the right tool for the job. I still think a 30 grain capacity case and a 6.5mm bullet is not the right combination for the job here but I would try the 140's before going to something I know will work like a 6.5 CM or 7.62.

LR-55

[grayfox]

I'm also of opinion that the grr is not quite up to the job...

For a physics problem I would make some simplifying assumptions, like you aim and hit the center of mass.
Then I should be able to calculate the force needed to tip a free-standing target over far enough that the downward (gravity) vector from the c.o.g. points outside of the base... that's in essence what causes an object to fall over or tip over.
So... mass of the target, rough dimensions, and front-back measurement of the base... then it's doable.
The ke remaining of the bullet as it strikes the ram at x distance, will translate into a force F=ma, so the question becomes how much force to tip the ram beyond its base... whatever units the various values come in, I will convert them to a single standard.
I think you're in the ball park. ft-lb/sec is a measure of power, like Kw, I'm thinking more on lines of energy and force.

I'm thinking the target is either 3/8 or 1/2" steel...
Will work on it more after my upcoming meetings in my consulting gig.

this thing has got me intrigued however...!

[grayfox]

So, is this ram, like, about 18" wide x 10" tall, or what...
I'll also assume the base (2-horizontal braces) is 6", 3" front and 3" back.