dustinflint says

More expensive to feed, but worth it.
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SteveD
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Re: dustinflint says

Post by SteveD »

I'm guessing that a Smallbore Ram weighs about 2.5 lbs. Using that assumption I calculated the percentage of the Smallbore Rams weight required to knock it over based on Jerry's measurements. Then I scaled it up to 55 pounds for HP Rams. Not at all what I expected.
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Re: dustinflint says

Post by SteveD »

Looks like this would work to measure the force it takes to push over a ram. I will find something and go do some measuring.

[img]
Mark%2010%20digital%20force%20gauge.jpg
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Re: dustinflint says

Post by thauglor »

Could solid core bullets (ie not lead) get around the problem?
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Re: dustinflint says

Post by Jerry G »

Remember Steve, Those numbers are not related to the energy you find on a computer print-out. It has to be a momentum of some sort but I left school a couple years ago and it escapes me. I'm thinking that I could look up the momentum (lb-f/s) on my Ballistic Explore program and say the largest number on the ram (body, front leg junction) is that and work the ratios back from there. My 500 M number is 40.3 which doesn't look like any other momentum number I have seen on other programs. It should work???? I'll have to set up a ram and shoot it with several other lighter bullets and prove or dis-prove the theory. There is another summer project. Retirement is fun.

I did it with a full size ram about 20 years ago and I have lost my data.
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Re: dustinflint says

Post by SteveD »

The following will address comments by both Jerry and thauglor.

First, thanks for your considered replies. :)

In parallel I have been talking to a manufacturer of solid bullets, thinking, as thauglor mentioned, that a solid bullet may be a solution by being able to transfer more energy to the Ram.

When I tried solid bullets 10 or 15 years ago, the technology was new and they had not yet worked out all the details. The ones I tried hit pressure limits well before I reached the velocities I could get with lead core/copper jacketed bullets of the same weight. The solid bullets made today have grooves so the the net length of the bullet that is of maximum diameter is spread out. This and perhaps a different alloy, allow you to get velocities similar to Lead/copper bullets of the same weight.

The other problem I saw was that the bullets shattered on impact, much like lead/copper bullets. Because of the pressure problems I gave up on the idea.

Fast forward to today. The guy that I have been talking to is familiar with Silhouette and the 'Ram problem'. He is also a mechanical engineer, as I suspect Jerry is. I asked him about the ram thing and about solid bullets shattering.

He was kind enough to work through the force question. Here is part of his reply (In blue.). I am not going to give his name as I have not asked his permission to do so.

"As far as staying together I know both the 105gr and 120gr will stay together nearly completely. I have always used an 8twist in 6.5mm and I think you'll be very happy with it. (That was in response to my telling him that I bought an 8 twist, 6.5 mm barrel yesterday.)

On the subject of tipping over rams I found that theoretically 13.75lbs of force should be enough to dislodge them. Now that seems REALLY light so I was vexed. Brings me to a new theory of momentum vs. energy.

Consider and 120gr 6.5mm bullet at a velocity of 2900fps it has 2240 lb/ft of energy but only 49lbs of momentum. this would surely knock over a ram (at point blank)

Same bullet launched at same velocity with a G1 BC of .510 should reach the target at 2000fps have 1050ft/lb of energy but only 34lbs of momentum.

Now we consider angle of attack of the bullet and realize the energy and momentum aren't perpendicular to the target and that half the energy an momentum will be deflected by the angle of force. Leaving us with 525ft/lb of energy but only 17lbs of momentum which is only slightly greater than the 14lbs needed to dislodge the target in a zero wind "all other factors zero" scenario."
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I play electrical engineer for my day job and am not familiar with the formula to calculate the force to tip the rams. Maybe someone else can run through it.

"Only 17 lbs" when 14lbs is needed seems to be a thin margin but it is more than 20% and this is with a 120 gr VLD.

I purchased the force gauge that I posted a pic of earlier in this thread. Once it arrives I Will get out to the range and take measurements on the HP rams and make a map similar to Jerry's.

I does look as though it will be an interesting summer project.
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Re: dustinflint says

Post by Jerry G »

I have talked to several shooters at the CBC last weekend and most agreed that a 6.5 mm 138 to 142 bullet traveling much over 2750 fps seem to blow apart on impact. Seems that they all think that those bullets traveling between 2600 and 2700 work just fine. I did shoot a 243 with a 7 twist bbl and pushed 107s at 2850. Low shots did not knock over the rams but the high ones did. A 142 at 2650 fps has quite a bit more recoil than the 107s for the other 3 but the 142s work just fine. Kathy shoots a 6 x 47 and she looses rams that are hit low. However she has a hold good enough she can hold high on them and they just keep falling over.

You gotta admit Steve, doing tests like this are fun and a good learning experience.
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Re: dustinflint says

Post by SteveD »

I wondered why you were not at the Smallbore match Sunday. :)

They are fun. You always find out something though not always what you were looking for. You don't even always know what it is you found out until years later.

I shot with Dean Hall Sunday. He was telling me that he had good luck with rams with a 123 Sierra out of a 260 Rem but not out of a 6.5 BR. Not sure what that means but it is a data point. Maybe the 260 Rem was above the minimum energy threshold but below the bullet burst threshold.
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Re: dustinflint says

Post by jask »

"Now we consider angle of attack of the bullet and realize the energy and momentum aren't perpendicular to the target and that half the energy an momentum will be deflected by the angle of force. Leaving us with 525ft/lb of energy but only 17lbs of momentum which is only slightly greater than the 14lbs needed to dislodge the target in a zero wind "all other factors zero" scenario"

Once again, we have someone stating the angle of impact on the target would take away half of the energy. That's ridiculous.
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Re: dustinflint says

Post by OldRanger »

2 things that seem to be missing. Angle of attack varies range to range. Ours is uphill, some are flat etc. also there is dwell time. Rounds that disintigrate have less time to impart their force. We need to shoot a force gauge. I wonder if I can talk someone at the university into lending me one.
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Re: dustinflint says

Post by SteveD »

Here is a model showing how to calculate the amount by which force is reduced when applied at an angle.
Force applied at an angle.gif
If F has been pushing down at angle θ on the object instead of pulling up at angle θ,


then the value for the normal would change to

N = mg + F sin θ.

Here is a link to where I got it.

http://dev.physicslab.org/document.aspx ... gangle.xml

Good point OldRanger, I was thinking about that today. At the range where I shoot the Rams are quite a bit higher than the shooting line so the angle of attack would be reduced. One of our shooters is a surveyor and measured all the animal lines. I will ask hi what the elevation to rams is.
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Re: dustinflint says

Post by OldRanger »

Yeah, basically if the angle is 45 degrees then the force is pushing half back and half down. I doubt we get that extreme of an angle, even with my .308. I would think more like 2-5 degrees (just a wag). 5 degrees reduces the force by somewhere around 6%. But as everyone has talked about, bullet composition has something to do with it too. It would be interesting to attach a small steel plate to a force gauge and shoot at it at the ram line with different bullets of the same weight to see the difference.

In terms of physics this is actually a very interesting problem. My old physics professor would have surely been out on the range with me if he were still alive. He came out to a hockey rink with me once to calculate the coefficient of friction on the ice before and after the zamboni ran on it. He just loved the pure science of it all.

That map Jerry made of the sb ram is very interesting. Thanks for doing that Jerry. If nothing else it shows us relatively speaking where it is harder vs easier and by what factor (assuming it scales).
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Re: dustinflint says

Post by Jerry G »

The angle the bullet strikes the target is very small. It has nothing to do with shooting up or down hill at the ranges we shoot at. Taken to an extreme, the angle up hill will make the bullet come into the ram at a more perpendicular angle. Shooting down hill would have a larger effect on that angle. Remember, if you are shooting a ram at the same elevation the angle of attack of the bullet will be down because of the bullet drop.

I am sure that the forces required would be proportional to the 1/5 size ram that I used. I am not sure how to calculate that constant factor but it does exist.
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Re: dustinflint says

Post by OldRanger »

Jerry, bring that force gauge to iron man. We'll weld a plate to it and do some shooting to get an idea how hard our bullets hit at ram distances.
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Re: dustinflint says

Post by SteveD »

200 yard zero, 500 yard total distance.
Just eyballing it the angle looks to be 35 to 40 degrees or so.

Increasing the launch angle as we do would reduce the angle.
308 bulet trajectory.jpg
I'm sure a 142 Sierra from a 260 Rem would be less.
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Re: dustinflint says

Post by xpilot »

"I can tell you that a 123gr bullet in a 6.5x47L is not going to do the job on Rams."

This is true...................why fight it. Pick the bullet that gives YOU the best success.
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