Ram-linel Trajectory Angle

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DanDeMan
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Ram-linel Trajectory Angle

Post by DanDeMan »

Ram-line Trajectory Angle

As some of you know, the issue of a bullet's terminal, trajectory-angle has been batted about for years. To finally drive a stake though the heart of that trajectory-angle bugaboo, let me share some test results from the spring of 05. The testing was done from the 1,000-yd line at the Sacramento Valley Shooting Center.

But first, a little background information gleaned from my early HP Silhouette days. Back then, my shooting pard, Tim Kurreck, was obsessed with his bullets' terminal-trajectory angle. Tim thought that slower, heavier bullets were not a good idea because he was convinced that excessive trajectory angle would make the rams harder to hit. His idea, which I heard from a number of top shooters back then, was that the height of the ram was diminished by the bullet's trajectory-angle. We had decided to develop a wildcat cartridge for HP what would have minimum recoil, but still get the job done on rams as well as have low wind deflection bullets to choose from. I wanted to go 7BR due to the availability of the 180 JLK's for rams. Tim thought that the trajectory-angle would be "mortar-like," and therefore make the rams harder to hit. So, I designed the 6.5TKS, which can launch the 139 Scenars to 2,775 fps.

Fast-forward to the early-to-middle of the last decade. BPCR Long-range Matches were my main focus; buffalo rifles fired from the 800, 900 and 1,000 yard-lines at Palma targets. I don't like recoil, so shooting a shoulder-stomping, large-case-capacity, 45-cal round was not where I wanted to go. So, I developed 38-cal rifles and bullets that would get the job done. But, the powers that be at the Whittington Center were convinced that my bullets would put the target pullers at risk of the perceived mortar-like trajectory angle of the 38-cal bullets. Several of us were shooting 38-70's that launched 370-gr bullets to about 1,400 fps, quite fast for a BPCR. Our bullets were hitting the Sacramento backstop much higher than the bullets fired from 45-90's and 45-100's. But, the "powers" at the Whittington center would not allow high, MV, 38-cal rifles to shoot the long-range nationals. Even when relating our experience at the Sacramento range, they refused to consider the issue.

So, a live-fire test was designed and executed from the 1,000-yd line at the Sacramento range to determine the terminal-trajectory angle. An article detailing the experiment was published in the summer 2005 issue of "The Black Powder Cartridge News." I'd link to it if it was online for those interested in the test details, but is not online. However, a couple of guys, after reading my article replicated the test at their range. Here's a link for those interested.

http://www.tcsa.info/HighPower/Bullet-Drop.pdf

One round I tested was a 45-70 that launched a 502-gr, Lyman, spitzer bullet to 1,227 fps with 70.0 grains of Swiss black powder. Most would think the bullet would be coming down at the target, when fired from the 1,000-yd line, like a mortar. The bullet's maximum trajectory height above the line-of-sight is about 35 feet, which occurs about 520 yards down-range, or 480 yards from the target. Most of you are probably thinking, "Man, that sucker is way up there so it should hit the target at a very steep trajectory angle." Not so!!! The average trajectory angle of the 10 test rounds was only 4.3 degrees below the horizontal.

The article linked to above has a nifty table on page 2 that lists their trajectory-angle results. Note that their 45-70 test-rifle, when fired from the 600-yd line, had a terminal trajectory-angle of only 1.36 degrees below the horizontal. And, the bullet only had an MV of 1,190 fps. So, the trajectory angle of even a 168-gr, 7mm SMK is probably less than 1/2 degree. Damn, need to run some computer simulations to get a better estimate...back in a Flash Jack...OK...Got a reasonable estimate. The 7mm SMK, when launched at 2,400 fps, has a 500-meter, terminal trajectory-angle of only 1/10th of a degree below the horizontal; so much for the trajectory-angle bugaboo.
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Dan Theodore
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Trent
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Re: Ram-linel Trajectory Angle

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1/10th of a degree??? So you are saying that the the 7br is a laser? That test that you linked to seems flawed. I am pretty sure that most of us would agree that a .22 LR fired at 200 yards isn't still rising. WTF? We've all watched through the spotting scope during smallbore matches where a round just misses over the back and impacts the berm behind the target, meaning that the ram blocks the view of the impact. I just don't buy it. No offense.

I have to leave to go pick up the wife at the Atlanta airport. I'll have to look back at this later. Maybe I am missing something.
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Re: Ram-linel Trajectory Angle

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The ground at the targets was 6 feet higher than the ground at the 200yd firing line where the rounds were fired from. When they say the bullets were still rising, it was because the shooters were aiming uphill at the targets that were six feet higher than their firing position plus whatever height the impact point was on the target (relative to the ground) above the guys shooting from prone (relative to the ground). That doesn't sound unreasonable.
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Re: Ram-linel Trajectory Angle

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Trent,

I wish I could link to my article written about this subject, but no dice. I don't even have an electronic copy anymore, crashed hard drive. Here's the section in the linked article that is of concern to you.

At 200 yd the bullet holes in the rear target were closer to the top than the ones in the front target. In other words, the bullets were still rising. This is because the soil protecting the target frames and pit crew is about 6 feet higher than the gravel located at the 200 yd line. The measured difference in bullet hole position for two bullets was -0.63” and -0.44”. At 300 yd the rear hole was lower on the target than the front hole. The measured difference for the only bullet we measured was 0.69”.

Since the target is about 6 feet higher due to the berm plus where ever the bullet went though the target, that might explain the rear target's hole being slightly higher than the front target's when the 22LR was fired from the 200-yd line.
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Re: Ram-linel Trajectory Angle

Post by Trent »

Yeah, I understand that point, and it's precisely why I call BS on the test. The only thing their test is good for is proving that their loads are safe at their range. My opinion.

Using my ballistics tables for a 7mm 168 Matchking at 2400 I am getting closer to 7/10ths of a degree at the Ram line. Seven times greater than your number.

I know one thing... one of us is wrong. :D

I've never been wrong by the way. Ever. :-B
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Re: Ram-linel Trajectory Angle

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Trent,

What procedure did you use to estimate terminal trajectory-angle?
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Re: Ram-linel Trajectory Angle

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I used a ballistics software for a 168gr SMK using Bryan Litz's recalculated BC for that bullet fired at 2400fps. I then used the formula for a triangle using the drop from 550yds to 600yds (21.9 inches). I know that's not perfect either, but there is no way in my opinion that the 7br is coming in at 0.1 degrees. That is basically flat. Although, 0.7 degree isn't exactly a steep angle either.
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Re: Ram-linel Trajectory Angle

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Trent,

Here's my process. First, 0.480 was used as the BC for 168-gr SMK launched at 2,400 fps. JBM's online, exterior ballistics software was used. I pluged-in the aforementioned parameters and set the the increment to 5 meters. The difference between 495 meters and 500 meters in drop was used to calculate the angle. I'll do it again to make sure I didn't make a mistake, of which I've been a pro at all my life. :lol: :lol: :lol: ....OK...I'mmmmm...Baaaack....here's the data as per JBM.

Meters --- Drop inches Edit for correction, those are MOA drops, not inch drops.
495 --------21.3
500 --------21.6

So, using high school trig we find the following:

Drop Delta - 21.6 - 21.3 = 0.3 inches drop from 495 meters to 500 meters

The distance for the drop is 5 meters, or (5 x 100) / 2.54 = 196.85 inches

Now we can get down to some trig.

Tan of terminal angle = 0.3" / 196.85"

The angle looks mighty close to 0.087 degrees

But, since 5 meters under-estimates the terminal-angle, I increased it to 0.1 degrees.

Slam...balls in your court dude.
Last edited by DanDeMan on Fri Apr 20, 2012 6:58 pm, edited 1 time in total.
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Dan Theodore
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Re: Ram-linel Trajectory Angle

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Trent,

I ran the same numbers as you for the average trajectory angle from 550 to 600 yards for the 168 SMK launched at 2,400 fps. The drop difference is 153.6 -125.3 = 31.3 inches of drop. The distance, in inches" is 50 x 36 = 1,800 inches. So, triging this out we get:

Tan (of average angle between 550 and 600 yards) = 28.3 / 1,800 = 0.90 degrees

Dude, it looks like I made a math error somewhere in my original calc. More sleuthing to come. OK, I'm doing the 495 to 500 meters calc again. Oh snit, I used MOA instead of inches of drop in the original calc. What a dunderhead!!! :lol: :lol: :lol:....OK, back to the calc:

Drop at 495 meters = 120.3"
Drop at 500 meters = 123.6"
Drop delta = 3.3"
Distance between drops = (5 x 100) / 2.54 = 196.85"
Tan (terminal angle) = 3.3 / 196.85 = 0.96 degrees

As you can see, the JBM site's software is not working well. I've seen this several times in the recent past, where results did not track with what we know.
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Dan Theodore
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Trent
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Re: Ram-linel Trajectory Angle

Post by Trent »

You're dividing the drop by the distance traveled to calculate the angle? I'll have to check that when I get home later tonight.
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Re: Ram-linel Trajectory Angle

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Trent wrote:You're dividing the drop by the distance traveled to calculate the angle? I'll have to check that when I get home later tonight.
Ya,

Divide drop by distance and then hit "Inv Tan" function. That will give the average angle over the distance.
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Re: Ram-linel Trajectory Angle

Post by Trent »

That doesn't seem like the right formula for this application. At best it seems more complicated that needed. Realistically this is a right triangle. We know legs a and b and can just calculate c, along with the angles.

0.1 degrees is just not realistic. As far as tranectories go, that is a laser. Hell, I was in denial with my number of 0.7 of a degree.
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Re: Ram-linel Trajectory Angle

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Trent wrote:That doesn't seem like the right formula for this application. At best it seems more complicated that needed. Realistically this is a right triangle. We know legs a and b and can just calculate c, along with the angles.

0.1 degrees is just not realistic. As far as tranectories go, that is a laser. Hell, I was in denial with my number of 0.7 of a degree.
Trent,

You did not read my mea culpa about my calc error of using MOA instead of inches for drop. Go back and reread my previous two posts.
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Dan Theodore
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Re: Ram-linel Trajectory Angle

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Ah! I missed that. I was just going to post that I ran your number using the right angle calculation and came up with .96.

Two roads... same destination. Haha.
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Re: Ram-linel Trajectory Angle

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Trent wrote:Ah! I missed that. I was just going to post that I ran your number using the right angle calculation and came up with .96.

Two roads... same destination. Haha.
Dude,

We were both wrong and right. :mrgreen: :mrgreen: :mrgreen: It's good to have fellow marksmen that are interested in understanding the effects of exterior ballistics and how said effects impact performance. Man, that is a bad pun. :mrgreen: :mrgreen: :mrgreen: ....impact performance...impact rams..Bwaaa, haaa, haaa...well, maybe not so funny after all. :lol: :lol: :lol:
Cheers,

Dan Theodore
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