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A decent spindrift formula

 
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  #1  
Old 11-08-2003, 11:41 AM
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A decent spindrift formula

A few of months ago one of the illuminate posted a spindrift formula as follows:

spindrift = (6/24.5)*(range/200)^2

His comment was that it was accurate +- 1/4 MOA. I've used it but am not happy with it and am looking for something a little more precise. Does anyone have any ideas about this?
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  #2  
Old 11-09-2003, 11:13 AM
MAX MAX is offline
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Re: A decent spindrift formula

Blaine, I'm moving tomorrow AM, books there, me here. You have McCoy's book I believe, perhaps there? If not, Robert Vaughn did some math in his book(Rifle Accuracy Facts) on the subject. Don't recall if it was user friendly stuff or not. I'll peruse their stuff when I get there.
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  #3  
Old 11-09-2003, 01:15 PM
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Re: A decent spindrift formula

Max,

McCoy does have a formula, but it is complex and uses variables not easily ascertainable. The rest of the well known works, including Vaughn, refuse to deal with spin drift in any meaningful way and just say that spin drift is too complex. In view of McCoy's formulas, I don't doubt the assertion, but I was hoping for something a little more precise than what I have.
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Old 11-09-2003, 06:54 PM
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Re: A decent spindrift formula

Theories on spin drift,

Please don’t criticize anything here as this is all theory and opinion and not necessarily fact.

Does anybody even have true calcs for spin drift?? Or known calculations? If you think about it SOME of the calcs would have to include the bullets stability factor and distance. Calcs for the stability factor includes atmospheric conditions, twist, velocity, and total bullet design, bullet construction and even the type of materials used.

If spin drift is a result of the bullet being angled slightly to the left or right depending on twist direction due to gyroscopic stability, or yaw of response, wouldn’t the bullets stability factor be necessary to give accurate calculations? If the bullet was being spun faster than necessary, wouldn’t that increase the spin drift? If the bullet didn’t spin, it wouldn’t move to the left or the right, right?

Something to think about.
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  #5  
Old 11-10-2003, 04:06 PM
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Re: A decent spindrift formula

meichele,
Your intuition is right on. McCoy has a nice chapter called "Linearized Swerving Motion of Rotationally Symmetric Projectiles" in which he explores spin drift in depth and sets forth the relevant equations.

One graph shows a comparison between the .50 cal, API, M8; the Match M118; and a Ball M80. At 2000 yards, the .50 cal has drifted 50 inches, the M118 about 95 inches and the M80 over 150 inches. Obviously the simple equation I have posted above doesn't do justice to the topic.
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