Spin drift question

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Don McDowell
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Re: Spin drift question

Post by Don McDowell »

Kenny no doubt reversals raise hell with things no wind zero or not
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Kenny Wasserburger
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Re: Spin drift question

Post by Kenny Wasserburger »

Jim,

I am betting you’re real close, with that.

Kenny Wasserburger


Come west shot in some wind, it will be Fun! They said. :lol:
We'll raise up our Glasses against Evil Forces, Singing, Whiskey for my men, Beer for my horses.

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Distant Thunder
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Re: Spin drift question

Post by Distant Thunder »

Kenny,
Thank you. I'm just sitting down to see what it will take to tip my tang sight 2.8 MOA at my 900 yard sight setting. This could take me a while to figure. If I can work it out it won't be perfect but it should be workable. I am most concerned about the long range end of things 800, 900 & 1000. The little bit it might be off a 600 on in closer should not be any worse than not addressing spin drift at all.
I never noticed any problems even at 600 yards, it's really at 900 and 1000 when I've scratched my head and wondered, " Why do I have 6 minutes left cranked in, it sure doesn't look like that much?". I was thinking that maybe I didn't have my mechanical zero setup right, but I do.

I'll see what I come up with and let you all know. Then I'll see if it works!
Jim Kluskens
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Re: Spin drift question

Post by Tomklinger »

Gentlemen,
I found a chapter on spindrift in an old book called Art of Shooting with the Rifle, printed in 1888. It says, for what it’s worth...
“When the fore-sight is adjusted correctly for the short distance, a Mark should be made. 3’ to the right of the short range mark to correct the drift of the bullet at long range, and this mark will be zero at 1000 yards for a rifle cut with a right handed spiral in the northern hemisphere in the temperate zone. With a left handed spiral this drift will only be two minutes of angle to the left. About four-fifths of this deflection is due to the spin of the bullet, which, in falling, gets more grip on the air below than above it, and rolls itself off to one side. The remainder is due to earth-rotation.
The lateral zero point for 600 yards will be half way between the two marks, that is, between the respective marks for short range and 1000 yards. I lay much stress on these positions being well ascertained, as, with what are called fish-tail winds, success depends on their accuracy. No wind blows steady from the same direction, and what in the main are head or rear winds will be found to vary incessantly to one side or the other, and must be watched and allowed for with the utmost care.”
Tom Klinger
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Distant Thunder
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Re: Spin drift question

Post by Distant Thunder »

Ok this is what I came up with so far. I needed to figure out how much shim I need under the right side of my sight base to move the rear aperture 2.8 MOA or .028" at my 900 yard sight setting.

I first set my sight to my 900 yard setting (roughly 2.000") and I measure as best I could from the tang up along the staff to an eyeballed center of the aperture, 3.950". Yea, I know but it's a Hepburn! That gave me 3.950/.500 = 7.900 (Height to CL/ Base width). Then I took .028/7.9 = .0035 (Move to the left/7.09 = Shim). So it looks like I need a .0035" shim under the right side of my sight base. I happened to remember that a pop can is somewhere about .003-.004". A quick check and as luck would have it a Coke can is .0035+.

Now I'll shim the base and get my mechanical zero reset at 100 yards by drifting the front. Then I'll set up at 100 yards with something similar to Don's cardboard and blue tape contraption to check just how many MOA it moves the POI at settings from 200 yards to 1000 yards.

More math :roll: , from my 200 yards setting to my 1000 yard is 130 MOA. From 100 yards to 1000 is roughly 143 MOA. That would require a piece of cardboard that is 13 to 14 feet tall! :shock: Is that right? That could get interesting. I think I'll have to go with some kind of sheet paneling or some such. I'll figure something out.
Jim Kluskens
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semtav
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Re: Spin drift question

Post by semtav »

I came across a bunch of papers I had misplaced. this was among them. Thought it was a timely find.





Dant spin Drift P1.jpg
DanT spin drift P2.jpg
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semtav
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Re: Spin drift question

Post by semtav »

So I looked in my Dan's Posts thread and came up with another;


Gents, Along with the silhouette-distance, windage-correction algorithm I posted on this thread, there is also the issue of the bullet's "spin drift" and its affects on perceived wind correction requirements. To refresh some memories, spin drift is caused by rapid bullet rotation; drift to the right from a right-hand-twist barrel and to the left for a left-hand-twist barrel. This phenomenon is not linear, it is a function of time-of-flight squared as well as the bullet's spin-rate (the faster the more the drift due to greater yaw of repose angle as spin-rate increases) and bullet length (the longer the more the drift due to more surface area to be acted on by differential pressure caused by the forward motion of the bullet and its yaw of repose angle.) I've heard it said that wind from 3 o'clock does not have as much affect on a bullet's trajectory as 9 o'clock wind with respect to horizontal displacement. When that is heard the thought that comes to mind is the speaker's rifle does not have its sights setup properly to cancel spin drift; that is, spin drift is adding to and subtracting to windage corrections. From 9 o'clock it is adding, from 3 o'clock it is subtracting. When we consider spin drift at silhouette distances, the distances in question in this thread, we should first look at horizontal displacement due to just spin drift at the c, p, t and r lines. To set the parameters for this discussion, assume that the rear sight-ladder is perpendicular to the front-sight bubble; that is, a "square" sight setup. And, let's further state that the load is cutting the "X" at 100-yds when the rear sight is set to mechanical windage-zero. Finally, HPT's load composed of the 45-cal, Paul Jones Creedmoor, 540gr bullet, launched at 1140 fps will be the target load, bad pun, eh...oh darn it's a double intender too. Oh, almost forgot, barrel twist is material, so HPT's twist-rate is specified to be an 18-twist. Next, a review of the effects of spin drift from 100-yds out to the c, p, t and r lines. At 200 meters, spin drift will horizontally displace the bullet about 0.4 MOA to the right from the 100-yd zero, rifle with "square" sight setup. At 300 meters, spin drift will horizontally displace the bullet about 0.8 MOA to the right from the 100-yd zero. At 385 meters, spin drift will horizontally displace the bullet about 1.1 MOA to the right from the 100-yd zero. At 500 meters, spin drift will horizontally displace the bullet about 1.6 MOA to the right from the 100-yd zero. As you can see, the above is not trivial, especially at the turkey and ram lines. To further get the point across, here's what wind corrections in a 10 mph 9 o'clock wind would look like for a rifle setup as described above. 10 MPH 9 o'clock Wind Chickens: 3.4 MOA Pigs: 5.8 MOA Turkeys: 7.1 MOA Rams: 8.6 MOA Now, compare that to a 10 MPH, 3 o'clock wind: 10 MPH 3 o'clock Wind Chickens: 2.6 MOA Pigs: 4.2 MOA Turkeys: 4.9 MOA Rams: 5.4 MOA As you can see, not setting your sights up properly can have a deleterious effect on your ability to score well in a rapidly switching wind condition. If the wind is fairly stable in direction one can just "chase the spotter", but when the winds start switching around the clock, far too many are toast if their sights are not setup properly. And, some rifles I've examined have had the rear staffs tilted to the right; that's even worse than having the staff perpendicular to the front bubble. We want our staffs tilted to the left to cancel spin drift so that our mechanical zero is our calm-wind zero at each yard-line we shoot at. Finally, it can be said; the further our targets, the more important proper sight setup becomes. The effect increases exponentially with range so at 1,000-yds we are talking 38.8" of right drift just from spin drift when shooting HPT's load. It is this writer's belief that a substantial amount of confusion arises from spin drift's effects on windage corrections. Setting one's sights to cancel this phenomenon might be the subject for future posts._________________Cheers....DanT I have no special talent. I am only passionately curious. Albert Einstein

Aviator
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Re: Spin drift question

Post by Aviator »

Distant Thunder wrote: Mon Feb 15, 2021 4:53 pm Ok this is what I came up with so far. I needed to figure out how much shim I need under the right side of my sight base to move the rear aperture 2.8 MOA or .028" at my 900 yard sight setting.

I first set my sight to my 900 yard setting (roughly 2.000") and I measure as best I could from the tang up along the staff to an eyeballed center of the aperture, 3.950". Yea, I know but it's a Hepburn! That gave me 3.950/.500 = 7.900 (Height to CL/ Base width). Then I took .028/7.9 = .0035 (Move to the left/7.09 = Shim). So it looks like I need a .0035" shim under the right side of my sight base. I happened to remember that a pop can is somewhere about .003-.004". A quick check and as luck would have it a Coke can is .0035+.

Now I'll shim the base and get my mechanical zero reset at 100 yards by drifting the front. Then I'll set up at 100 yards with something similar to Don's cardboard and blue tape contraption to check just how many MOA it moves the POI at settings from 200 yards to 1000 yards.

More math :roll: , from my 200 yards setting to my 1000 yard is 130 MOA. From 100 yards to 1000 is roughly 143 MOA. That would require a piece of cardboard that is 13 to 14 feet tall! :shock: Is that right? That could get interesting. I think I'll have to go with some kind of sheet paneling or some such. I'll figure something out.
Jim, just thinking out loud here.
I do not think that the 3.950 distance from the 900 yard setting center of aperture to the tang is the relevant dimension. I think that the .028 offset needs to be from the 900 yard setting center of aperture, to the 100 yard setting center of aperture.
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Distant Thunder
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Re: Spin drift question

Post by Distant Thunder »

Steve,

I see what you are saying and I need to think on it some. That would seem to be a lot more tilt to the left, correct? I mean if you worked from the 100 yard setting. More than twice as much actually. You'd be tilting the sight .028" in 1.430" of the height instead of tilting it .028" in 3.950".

What you're saying sounds correct I just need to think that through and it's getting bit late to do that clearly tonight. Tomorrow morning after a couple of cups of good strong black coffee my brain will be as sharps as it gets!

if I get that staff leaning over too far I'll be shooting around corners for crying out loud!

I have not actually done anything with my sight yet I didn't see doing it while I can't shoot the rifle to confirm any of my numbers. It needs to warm up a bit. It was 18 below here this morning and it slooooooooowly warmed up to about 5 above late in the day. It was a good day for pushing numbers around on a piece of paper!
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MikeT
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Re: Spin drift question

Post by MikeT »

Has anyone used the 12.5 yard distance to determine where your rifle shoots when set at a mechanical zero windage? You can see how much your rifle POI changes based on where you set the elevation for 800 to 1K.

Keep on hav'n fun!
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Distant Thunder
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Re: Spin drift question

Post by Distant Thunder »

Steve,

I wasn't going to sleep with this on my mind anyway! Thank you!

Chew on these numbers and see what you think.

Per bruce, the spin drift difference between 100 yards and 900 yards is 2.5 MOA, 2.8 at 1000 minus the .3 at 100 = 2.5. Then the difference between my 100 yard setting and my 900 yard setting is 1.230". I need the rear sight aperture to move .025" to the left in 1.230" of elevation. So 1.230/.500 (base width) = 2.460 and .025"/ 2.460 = .0101". Now if that is correct then I need a .010" shim under the right side of my base.

Oddly enough the last time I sat down and made a guess at what shim size to use I came up with .010", but I have no idea how I arrived at that number, but I had gone and dug up a sheet of .010 stock to make a shim and it's been sitting on my bench since last fall. I must have notes somewhere?

This sound correct I think, but my staff is going to be leaning way left a noticeable amount, .010" per 1/2 of height. The good thing about this is anyone could add a .010 shim and be in the ballpark since this is only an approximate correction for Creedmoor rifles. The numbers between 100 and 900 should be fairly close with any reasonable load. As you pointed out that's the elevation change we are dealing with and not the height from the base as I was thinking.

That's it for now, good night!
Jim Kluskens
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desert deuce
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Re: Spin drift question

Post by desert deuce »

I just took an aluminum Coke can, a regular pair of scissors and cut a shim and placed under the right edge of the rear sight base on my Hepburn.

No micrometer, no chalk and no axe.

Why? Because that is what Dan'l told me would work. So I did it. Shim is still there. Guess it works OK.
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Re: Spin drift question

Post by bruce m »

and there was i thinking you were the master of the windage foresight all this time.
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Re: Spin drift question

Post by bruce m »

and there was i thinking you were the master of the windage foresight all this time.
the sight adjustment for dans 38 cal would be somewhat greater than the ave 45 cal.
this is due to the 38 having a 10 or 12 twist, and the 45 having around an 18 twist.
you really need to work out the spindrift for your load in your rifle before spending time and effort working out how to deal with it.
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Distant Thunder
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Re: Spin drift question

Post by Distant Thunder »

Why do I feel like dog chasing his tail? :?

I love this ballistics stuff, it fascinates me! I don't have a clue, but I'm sure I'll have a hold of it if I just go around one more time....... :lol: :lol: :lol:
Jim Kluskens
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