Range Report Spindrift Calculation??

Re: Spindrift Calculation??

Hey just shoot a bit of 308 1000 yard F TR Class these days, but I do put on 1 moa left at 1000 yards, .75 moa for 900, .75 moa for 800 and .5 moa for 600 yards, in no wind conditions for spin drift. I do not know if all rifles exhibit it but mine seem to do with 155s at 3000fps in 11 twist bbls.

In old days shooting at just steel or human sized targets I did not do this but since I now shoot at a .50 moa X rings it seems to make slight difference so I do it. Now shpoting on steel or human sized taregt I just hold slightly and I mean slightly left for same effect
 
Re: Spindrift Calculation??

Trigger,

Great explanation.

Why anyone would advocate ignoring -in error budget terms - an absolute 'known' is beyond me
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For this discussion to go anywhere, it'd have to go back to first principles to get everyone 'on net' and understanding the ballistic concepts behind (spin)drift....clearly some here don't actually believe(spin) drift exists whilst others believe (spin)drift values vary as a result of local wind changes...and are unable to conceptualise it as a seperate effect.
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I've given up on this thread; I'll watch with interest to see how much luck you have in convincing them!

Good luck!
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Re: Spindrift Calculation??

Admitted, i'm anal retentive on precision... 1500+ meters for 95% of my last 12 shooting years has beaten into my head, DON'T IGNORE NOTHING.... kind of like the Rangers 1st rule, Don't Forget Nuthin... 5 Degrees of air temp change in all but the highest performance bullets means a miss, just like the 1-2 MPH wind variable.

 
Re: Spindrift Calculation??

....oh........thanks for the brain damage guys.......Chad you bonehead....if you want to know why you missed the 1000yd shot at tacpro......call me ....you missed one subtile nuance that is only exibited on that range.......maybe, just maybe all bets are off though because of the new steel mover and all the adjacent construction......


[email protected]
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Tactical</div><div class="ubbcode-body">I do not know if all rifles exhibit it but mine seem to do with 155s at 3000fps in 11 twist bbls.

In old days shooting at just steel or human sized targets I did not do this but since I now shoot at a .50 moa X rings it seems to make slight difference so I do it. Now shpoting on steel or human sized taregt I just hold slightly and I mean slightly left for same effect </div></div>

My .308win and a few other 308 win's I shoot with seem to agree with your findings as well, and five of those guys are in your old, line of work.
My 300wm, throwing 190's at 3125-3140 out a 10 twist shows 5/8 moa left.
I do much better on both of jhuskey's "Herman's" than his target plates,
but I don't know why?
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Re: Spindrift Calculation??

Again, nobody is saying it doesn't exist, however, even by your own admission, that shooting styles, and "hot rod" rifles can change it, the actual number is unverifiable to say what you are using and my contentions is, shooting in a no wind situation is few and far between as being realistic. Even if you don't feel wind at the shooter, you still have a 1000 feet of ground or more to cover and you have what is above you as well. (Anyone who has shot on a tower can tell you it is night and day difference between the ground and 1 flight up where the bullet actually lives.)

Based on my experience you have everything from .6 MOA to 1.2 MOA of potential drift as per the resident experts on it. That is a pretty wide spread to stake your case on. As well, if the numbers were verifable to any degree even with a wind correction you still have to contend a 1.2 MOA error is a miss.... yet people aren't missing... and many hit center just fine.

The missing 1 MOA or what I believe is a number twice rounded up, would work both ways for anyone shooting to 1000 yards. And since most people shoot a target 18" or smaller, why are wind calls working.

Basically the contention is, the math doesn't add up, and the difference it bigger than the benefits. Explain to me how anyone leaving out a 1.2 MOA correction can hit a target at 1000 yards... Luck ? And why does a persons shooting style affect the downrange outcome.
 
Re: Spindrift Calculation??

It's funny you should mention that, because in doing some research on this subject, the best information I have found comes from someone outside the shooting world who is looking at spin drift much in the same way a baseball player throws a curveball.

To paraphrase, and I liked his analogy, the curve ball is highly dependent on the laces, to which he explained that spin drift is effected by the grooves in the bullet cause by rifling. Basically spin drift would be nearly eliminated according to this physicist if the bullet remained smooth.

So, using the idea that the grooves are what is is actually causing drift, it would venture to say the depth, size and number of grooves would effect the amount to a large degree, not to mention speed and surface area of each bullet. So it's safe to say, there is no one number as all things being equal, they are not as no two rifles would actually exhibit the same amount of drift unless they both equally affected the bullet in the exact same way. number or grooves and well as depth, and since the grooves of a bullet is almost like a fingerprint...

see my point.
 
Re: Spindrift Calculation??

actually a ratchet cut barrel mostly eliminates the groovey part of the boolet ...at least on the opposite of the driven side & thus cuts down on the air friction.....in theory of course...
 
Re: Spindrift Calculation??

Lowlight 1+

Dean I have the utmost respect for you and your ballistic knowledge. I don’t think anyone is saying spin drift doesn’t exist. Spin drift is an anomaly that exists in shooting and one that cannot be accurately calculated given all of the variables in shooting long ranges. About the only thing you can truly do is compensate for it.

Too many people (arm chair experts) here jumped on this SD band wagon without understanding the bigger picture and are the first to tell you how wrong you are for not correcting for it. I can’t say I’ve ever heard or will ever hear Bill at TAC PRO, or Rod Ryan at Storm Mountain telling folks on the 1000 yard range or beyond to "Remember to correct for spin drift".

I will also never make a claim that I shoot beyond 1100 yards on a regular basis. I have shot over 1500 yards on several occasions with some outstanding .50 cal shooters and I learned there were always a few constants at shooting at those ranges. .

First there is no such thing as a calm day…EVER, they don’t exist! Second, the wind never blows at a constant speed and direction over the entire bullet flight path. Third there are always crosswinds at different directions, angles and different speeds. Fourth, on a 3/4 man size steel silhouette target there was almost never a 1st round cold bore hit at 1500 yards. Actually we had only one and that was a great call, spin drift was the farthest thing from most of the shooters minds, figuring out the winds the biggest challenge.

If there was a hard and fast mathematical calculation for spin drift everyone would be using it and ballistic programs would have it for each bullet and bullet type. But as I’ve said at the end of the day there are too many variables to accurately calculate for spin drift. You can say you compensate for it, but are you really compensating for spin drift or for wind or both as one? I’ve never heard a shooter or a spotter saying I need x amount clicks right or left for spin drift and on top on that I need x amount for wind.

For me and everyone I’ve ever shot with, we compensate for any spin drift that exists in our windage corrections. Say what you want at those really long ranges there are just too many variables making just a specific spin drift correction almost impossible.
 
Re: Spindrift Calculation??

This is the technology behind CheyTac controlled spin that slows the rotation allowing the bullet to nose down during flight for less "keyhole" flight
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">It's funny you should mention that, because in doing some research on this subject, the best information I have found comes from someone outside the shooting world who is looking at spin drift much in the same way a baseball player throws a curveball.

To paraphrase, and I liked his analogy, the curve ball is highly dependent on the laces, to which he explained that spin drift is effected by the grooves in the bullet cause by rifling. Basically spin drift would be nearly eliminated according to this physicist if the bullet remained smooth.

So, using the idea that the grooves are what is is actually causing drift, it would venture to say the depth, size and number of grooves would effect the amount to a large degree, not to mention speed and surface area of each bullet. So it's safe to say, there is no one number as all things being equal, they are not as no two rifles would actually exhibit the same amount of drift unless they both equally affected the bullet in the exact same way. number or grooves and well as depth, and since the grooves of a bullet is almost like a fingerprint...

see my point. </div></div>
 
Re: Spindrift Calculation??

22 pages of this?

Here is how I calculate...I went to the range yesterday. And last weekend. And a few days before that. I will admit that I had a two week stop when I was finishing up my classes.

And, when I finially got on the Grand, balistic calculator math was incorrect. Oh, but it says!!!!! BFD. How many of you are really going to rely on this type of math? Not me. It may get me in the ballpark, but my range book will do the rest. .19 MOA at 1000 yards? Thats two inches. I doubt very seriously that too many on here can perdict a two inch drift.

Dope it, shoot it, dope it again, and shut the hell up.

I have more bullets down this one gun in two months of getting it back, than many have before they hock their is the forsale section.
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">Again, nobody is saying it doesn't exist, however, even by your own admission, that shooting styles, and "hot rod" rifles can change it, the actual number is unverifiable to say what you are using and my contentions is, shooting in a no wind situation is few and far between as being realistic. Even if you don't feel wind at the shooter, you still have a 1000 feet of ground or more to cover and you have what is above you as well. (Anyone who has shot on a tower can tell you it is night and day difference between the ground and 1 flight up where the bullet actually lives.) </div></div>

I'm not saying the S/D <span style="font-weight: bold">not </span> added will cause a miss at all, what I see happening is with the S/D not added I am always adding more left than anything else. What I might read and call a 3mph FV you might call 2mph FV even though both sticks are the same/same except the cant I have put in my glass. I believe the larger the pill, the more s/d, also the slower the pill the more s/d as both build air pressure differently. Hence the .308 10 twist 175smk at 2725 vs the 10 twist 300wm 190 smk at 3125-3140. When I have the 300 rebarreled I'll know if just the T/R makes a diff., as the new one will be 11.27. Also this next H/R the new .308 stick is 11.27 so I'll be looking at wind calls vs the old 10 twist as well.

Again I don't see S/D causing a miss because most folks dial for what <span style="font-weight: bold">they are seeing/calling </span> and doping based on <span style="font-weight: bold">past data </span> . If one does not have history to go by, then it matters not if they add 2 moa or 3moa wind as long as their hits are in the zone.

Not sure what works for everyone else but, when Dtubbs explained it to me, I tried it and found it just works for me, may not work for anyone else on this rock but him and I, don't know.

I have shot to 1k with no wind (via smoke pots) and with the left canted retical they both hit dead ctr. Could have been the moons phase interlocking with the northern lights, based on ice thickness...could be?
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Re: Spindrift Calculation??

Gunfighter, I hate to say it, cause I know what I sound like, but you are discounting a whole lot of math.

If one uses something as simple as 1 MOA to address SD, on any 18" wide target or less you have a miss... there is no way to explain it away.

If I kestrel the wind, and input the data from any program I use, whether it be Patagonia, Horus, Exbal, FFS, if that gives me a solution and I make a first round hit, which generally happens for me, something is deeply wrong. There is no way I could have made a 1 MPH error every time I hit, because that is essentially what you are saying. An Error of 2 MPH won't work, an error of being dead on with the wind won't work, only being off 1 MPH will. You can't fudge the math.

So either people are extremely lucky, and the programs are wrong in their premise of assisting the shooter in first round hits, along with every single wind table ever published, because there is no physical way to hit a target center with the numbers put forth... so there has to be another explanation, like the number for SD must be smaller. and in reality with the majority of right hand shooters, the problem should be two fold, by combining with trigger errors.
 
Re: Spindrift Calculation??

The 2 vs 3mph was just numbers thrown, but I do believe everyone reads mother differently and I think this is where the difference comes in. What I was told/taught as 1-3 mph does not run with want Serria says the bullet does based on my target.

Math will always render the correct answer, assumeing correct input. I'm not sure where the end if this debate is, but I know something is happening other than wind down range, and the sd cant I use seems to work for me.

If everyone read/ranged everything the same, no one would never miss. All the programs seem to arrive at about the same answers only after the same field data is inputed.
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Gunfighter14e2</div><div class="ubbcode-body">

Math will always render the correct answer, assumeing correct input. I'm not sure where the end if this debate is, but I know something is happening other than wind down range, and the sd cant I use seems to work for me.

If everyone read/ranged everything the same, no one would never miss. All the programs seem to arrive at about the same answers only after the same field data is inputed. </div></div>

I know this is completely beating a dead horse, but I think you are getting closer to seeing things a little differently, but since we're here, I'll simplify the math a bit.

1 MOA @ 1000 yards is 10 inches

Doppler says 175gr bullet has 11.43" of Spin Drift @ 1000 yards, which is now being read as a verifable number by some... in fact an "ah ha" moment in this discussion.

So, in no wind situation a center hold on an 18" wide plate means only one thing -- the shooter misses the plate each and every time by 1 MOA or in the case of doppler 2.43" to the right.

Now, add that to each and every wind call and automatically the shooter misses the plate by the same numbers, there is no way to use any wind table to any effect without the addition of pure luck because the shooter will always have to be either 2.43" short or long on their call, period. Cause without that additional correction to center you still miss the plate.

The math gets no simpler than that, unless there is something else affected by select shooters to influence this number, which in my contention is trigger control or some element of the fundamentals of marksmanship, which I say is more common and greater a factor.

So to say say spin drift effects every 308 rifle by 1 MOA or more is either wrong, or more so, X number of shooters pull their shots right more than Y accounting for the math not working out. In which case, X adds in spin drift rather than dialing for "their" trigger or flinch.

Is it the gun, or the shooter ? cause if its the gun, my math is undeniable.
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there is no way to hit the target without the correction wind or not.

Now, finally so there is no misunderstanding me, I contend SD is equal to .5 MOA or less depending on the rifle, but in essence half of the noted number at 1000 yards with a 308. Other bullets and calibers have other numbers and no two rifles are exactly the same.
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Re: Spindrift Calculation??

I understand your math, and think neither will convert the other, based on what we both see as the final product.

I still believe the difference is in the read before taking the shot.

The USO thats on the 300wm is 1/2 IPHY both ways, (0.497 moa) now using your math, I should never hit ctr on a 1moa target at 1k w/o up to 1/2 IPHY hold off?
Again the read was it 2 mph, 2.5 mph or 3mph
 
Re: Spindrift Calculation??

I'm afraid that anyone who believes that the effect of spindrift on a 175SMK at 1000 yards is 11.43 inches needs a lesson in experimental error.

There is no way that number can be known to the precision of two decimal places. I'm suspicious of the claim that it's been accurately measured to an accuracy of several inches, unless that doppler radar test was done in an enclosed building where there could be no possible wind effect - which it wasn't - or the figure was the statistical result of many shots, which I question. Even so, anyone who believes that result to 2 decimal places is smoking crack or is innumerate.

Assume, though, for the sake of discussion, that the effect of spindrift on a 175 SMK at 1000 yards is one MOA.

So, dial it, or hold it, and see what happens.

However, Dean listed above 7 possible sources of horizontal error.

In the category of shooter induced error, there are actual several: failure to hold the rifle straight up, failure to have installed the scope so that the reticle is vertical, jerking the trigger, etc.

In addition, there is the possibility that the erector system in the scope doesn't track straight up, even if the reticle is vertical. I've seen scopes where the horizontal and vertical reticles were not orthogonal, i.e., at right angles to each other. When you have that, which one is incorrect with respect to the elevation erector system?

I have launched hundreds of 175 SMKs at one particular 1000 yard target using a rifle with a right-hand twist in the barrel.

I have far more frequently missed that target to the left than to the right, in a ratio of probably 10 to 1. (And, no, I don't always miss.)

Should I conclude that spin drift is moving my bullet to the left?

I think not. The prevailing wind on that range comes from about 135 degrees relative to the target - and there's almost always some.

Assuming that the effect of spin drift is about the same magnitude as that of a 1 mph full-value crosswind at distance, I am similarly highly suspicious of the ability of any individual or system to measure the wind over a 1000 yard or longer range to an accuracy of 1 mph.

So, now, when I shoot at that target when I think that there is no effective wind, I hold a bit to the right - and mostly that works. Is that the effect of the wind? Spindrift less the wind? Is my rifle canted? My scope? I don't know. I don't care.

If I miss, there's a much higher probability that the second round will hit. Good enough for me.





 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Quote:</div><div class="ubbcode-body">If I miss, there's a much higher probability that the second round will hit. Good enough for me.
</div></div>

OK now it is complete. Lindy has spoken.
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This topic has some good points. If you believe that spin drift is there then adjust for it. I would think that wind would be something that I would worry about.
 
Re: Spindrift Calculation??

Come Pete, it's not even close to complete, I have a whole new line of "shooter error" to explore.

Here are some great examples why I have to say SD is lost in the noise, note the targets shot at 100 yards and where the majority of shots have impacted... SD or SE, you be the judge:

SHGroupTarget_small.jpg


How many right of center and what do we think that means downrange ?

223_1_5Medium.jpg


More right of center.... SD or SE

308_1_5Medium.jpg


I like this one above because one group goes left.

I think demonstrates the point pretty effectively.

 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lindy</div><div class="ubbcode-body">I'm afraid that anyone who believes that the effect of spindrift on a 175SMK at 1000 yards is 11.43 inches needs a lesson in experimental error.
</div></div>

No, they need a lesson in ballistics
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.

The same 175smk will have different (spin)drift depending on how it is fired. Different MV and spin (ie gyroscopic steadiness)will produce different (spin)drift in the same projectile.

Similarly, I note the schoolboy concept of 'it's the grooves in the bullet' has crept in.

Fellows, anything that you read on the mechanism causing (spin) drift that doesn't include the word 'precession' is false science.

All this mutual backslapping by those applying schoolboy concepts to 'prove' it doesn't exist or think (spin)drift has a fixed value for a given bullet is so repeatedly ballistically ignorant it's laughable.

For the purposes of an internet forum discussion; this fellow sums up the bottom line well:

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: SFC Carpentier jr</div><div class="ubbcode-body">Again if you feel that spin drift needs to be adjusted for , DO IT, if not F$%k it. </div></div>

 
Re: Spindrift Calculation??

Lowlight,

I'm still trying to understand the math?
Throw out SD on the 18" plate and insert a leupold M3.

I'm not trying to make light of anything, just trying to understand the concept.

 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Gunfighter14e2</div><div class="ubbcode-body">Lowlight,

I'm still trying to understand the math?
Throw out SD on the 18" plate and insert a leupold M3.

I'm not trying to make light of anything, just trying to understand the concept.

</div></div>

Forget it... a scope type has nothing to do with anything, if you can't figure out what I wrote, I can't help you. canted scope or not.
 
Re: Spindrift Calculation??

I dont know why guys are getting wrapped around axle on this. Long and short from my way of looking is.

On Human sized target to 1000 yards no bother because wind misread will be greater and it does get lost in noise.

For precision paper shooting and Extreme Range (past 1000 yards) you best figure it in to increase your already crappy chances of first round hit.

I went over my log books for last couple of years shooting 1000 yard comps. I found a mix of wind corrections right and left. I found where I read left to right wind as being equal to right to left wind but corretion was not equal. It always seems to take more wind correction for left to right than right to left for me and my weapon. Now this is not perfect data by a long ways. But enough for me to notice.

Its not reticule cant because I do simple box testing scope and know my adjustments are true up and down

Its not trigger control because my scores have been too consistent.

Now what Lowlight mentioned of inducing purpose error in initial sighting is something that would work as long as you did not do HRT work( where you need POA to be POI You would be screwed in court attempting to explain why you zero'd 1/2MOA left and missed bad guy or worse hit hostage. I only bring this up because some guy will think Lowlight was serious) and your bullet had constant drift amount.

ON Baseball spinning because of laces. I have thrown enough baseball and smooth balls, and induced curve to say its not just the laces. Its the spin and even a smooth ball can be made to curve a great deal. Coem on I am not the only one to pitch before? Whiffle balls, Ping Pong Balls you name it us poor kids played baseball with them.

Now brings up a question. Are all bullets and twists equal in drift? I dont believe so. I wish someone had hard data to look at on this.

On Badland Parlor trick. No trick. It works same way as launching bullet over target and hitting distant target. Except in thier case they launch to side of target and let wind correction bring bullet back in align with distant target after clearing blocking target. Drill is designed to teah folks to know path of bullet and be able to avoid obsticles. Not alot of practical application but makes a man think just the same. Short version is place one target at 100 yard. tehn place another target 600 yards directly behind it. Dial in 3 moa of left wind. Hold 6 moa right of target, Fire. Bullet passes by first target missing it right by roughly 3 moa. The round continues to drift to left because of sighting error and hits rear target 12 inches left of right edge. Not a trick at all.
 
Re: Spindrift Calculation??

This thread is a close second to the coriolis thread as the best thread ever that should be stickied for all eternity.

I'm waiting with baited breath for the thread on the gravitational effects of the moon throughout its different phases on bullet ballistics. Damn, a guy can dream.
 
Re: Spindrift Calculation??

<span style="font-style: italic">Well, not wanting to let it end I was contemplating something in the shower this morning, it was either spin drift or conditioner, and I chose spin drift. </span>

I pictured this:

Shooter "A" walks on a range one sunny afternoon and finds another shooter on the line. Asking how it goes on this beautiful day, shooter "B" says, <span style="font-style: italic">"you know on this perfect day where the wind flags aren't moving I was hitting to the right a lot at 1000 yards".</span>

Shooter "A" says, <span style="font-style: italic">"well, when I talking to DTubb he explained to me about spin drift and now I dial 1 MOA left on my scope and my hits are much better. See the bullet is spinning to the right and you have to compensate because at a 1000 yards doppler radar says you have 11.43" of drift with a 308 175gr".</span>

Shooter "B" says. <span style="font-style: italic">"wow, I didnt know that, thanks, that is great I am gonna try that so my hits are better too". Shooter "A" knows what he is talking about".</span>

Or you can have the same exact situation above and Shooter "A" can say, <span style="font-style: italic">"well, it's possible there is a small amount of wind between you and the 1000 yard target, or quite possibly that $1200 Leupold scope you bought has a canted reticle which is cause you to impact right. And I see you are right handed shooter, I have to say it is quite possible your trigger control is lacking pulling your shot right. On top of that, I see the groups you shot at 100 yards are favoring right of center as much as .5 MOA... I mean there are lot of things going on here, or even a combination of small errors on your or your rifles part is causing you to miss right".</span>

Shooter "B" says, <span style="font-style: italic">"what do you mean my $1200 Leupold might be off, and my trigger control is fine who are you? You weren't here to see me pull the trigger and the wind flag at 1000 yards isn't moving, so there can't be any wind. what errors am I making, you just walked up here and I never met you before".</span>


<span style="font-weight: bold">My point,</span> one answer makes you look pretty damn smart, the other makes you an asshole.
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carry on...
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body"> hear </div></div>
here
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Captain Kick-Ass</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body"> hear </div></div>
here</div></div>

where did I put "hear" I can't find it...
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">
Shooter "B" says, <span style="font-style: italic">"what do you mean my $1200 Leupold might be off, and my trigger control is fine who are you? You weren't <span style="color: #FF0000">hear </span> to see me pull the trigger and the wind flag at 1000 yards isn't moving, so there can't be any wind. what errors am I making, you just walked up here and I never met you before".</span>


<span style="font-weight: bold">My point,</span> one answer makes you look pretty damn smart, the other makes you an asshole.
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carry on... </div></div>

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Re: Spindrift Calculation??

Lowlight <span style="font-weight: bold">My point said:
<<GRAEMLIN_URL>>/smile.gif[/img]

carry on...

OK- I jumped out of the ballistic software thread and withheld from input here --- till now.

Spindrift Calculation-- for right hand twist barrels may require you to lean to the left prior to firing. When you fire the bullet will curve right into the X-ring. Matter of fact I even seen angelina jolie teach other people about it in her new movie (I saw her do it only a couple times though because I had trouble taking my eyes of her guns!)

Ballistic Data Card says "for shots over 600 meters- <span style="font-style: italic">L E A N ! !</span>

Please don't reply if my answer makes me look like an ashhole- I like to think it makes me look Smart !!

By the way I got my idea watchin angelina, I won't talk about my thoughts in the shower after that.
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Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Captain Kick-Ass</div><div class="ubbcode-body">"You weren't hear to see me pull the trigger"


el nino perhaps?</div></div>

I'm a dumb ass... thanks for the catch, totally negates everything I said now... disregard.
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Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body"> because at a 1000 yards doppler radar says you have 11.43" of drift with a 308 175gr". </div></div>

Lowlight; are you actually reading (understanding
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) anyone else's posts?
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I say again; it is not possible to define (spin)drift for a particulary calibre and bullet weight.....additional data is required from which to calculate the bullet's stability.

[It is, of course, possible to define it for a certain issue weapon type firing a certain issue ammunition]


I have to say, that for someone stating 'no one's denying (spin)drift exists' your analogies sound pretty much like someone who's trying to find anything to say it doesn't
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Given your earlier clutch at a 'grooves on the bullet' understanding of (spin)drift; may I humbly suggest that you do some reading of artillery texts on the subject ...this is all fully quantifiable, calculatable and old news to Gunners....have a look at the formulae I posted on page 9 or 10 or 11
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err..page 8
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Re: Spindrift Calculation??

Brown Dog, go read Triggerfifty's post, he stated in 2001 doppler radar was used to track spin drift of a 175gr out of a M24....

Maybe you need to read more.

Artillery and small arms have very little in common... trying to use artillery standards is what fucked mils up...

go back and read what was posted and what I said, you are the one who doesn't get it.
 
Re: Spindrift Calculation??

Not interested in any pissing contest so my interest here is strickly learning. Even an old dog can learn or they just lay down and die.

What I would like is someone with some true knowledge on this to tell us what proven facts are. Is there a method that we can use to calculate drift based on bullet diameter, weight, and bbl twist rate. Velocity would probably factor into this also.

I just have data from shooting the distances and bullets and would love to see something more on this.

It would be nice to have for extreme range shooting. I am happy with how I do the out to 1000 yards but would like to play at 2000 yards some more.
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: jwp475</div><div class="ubbcode-body">

Maybe brown Dog will calculate the spindrift for the various bullets and calibers that we use. A very nice gesture indeed.</div></div>

Exactly, the math is there, Brown Dog, calculate the drift for us and let us know what happens when you change "miles" to "yards"

I wonder why the military never employed artillery gunners as snipers, they know stuff the rest of us don't.
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">Brown Dog, go read Triggerfifty's post, he stated in 2001 doppler radar was used to track spin drift of a 175gr out of a M24....

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and as such it will only describe the (spin) drift out for that 175gr out of an M24
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and not all 308 175gr bullets out of any rifle
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Which is why I said

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Brown Dog
I say again; it is not possible to define (spin)drift for a particulary calibre and bullet weight.....additional data is required from which to calculate the bullet's stability.

[It is, of course, possible to define it for a certain issue weapon type firing a certain issue ammunition</div><div class="ubbcode-body">


</div></div>
 
Re: Spindrift Calculation??

As far as judging small arms fire with artillery standards ??
... if I had a dollar everytime someone told me I missed by a mile !!

Seems like it would come down to simple let's see it be done kind of thing. Once we all believe then we can begin to understand- the founding principal of any religion, shooting included.

Anyone have a left hand twist rifle barrel they can load up with the same ammunition and fire at the same time next to the right hand twister???

Would that be one of those results speak for themselves kind of things. I am guessing that at 1K the right twist will hit to the right of the X and the left twister will hit to the left. Anyone hear of this comparison being done?

Once we SEE that spindrift does occur- and believe what we just saw then the game begins in how much and at what distance it occurs with our particular load/rifle combo.

This opens it up for the experts to predict it for us because we all believe it occurs.

Left twist VS. Right twist-- anyone ???
 
Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">
Artillery and small arms have very little in common... </div></div>

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My friend, a spin stabilised projectile in flight is a spin stabilised projectile in flight --they have EVERYTHING in common
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Re: Spindrift Calculation??

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Tactical</div><div class="ubbcode-body">
What I would like is someone with some true knowledge on this to tell us what proven facts are. Is there a method that we can use to calculate drift based on bullet diameter, weight, and bbl twist rate. Velocity would probably factor into this also.
</div></div>

Apologies, I'm working through the torrent of abuse in bite size chunks!
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Mike,

Stability calculations require data gathered from firings (such as the dreaded Doppler!)....and they will only apply to that proj at that mv at that spin rate.

JWP, LL,

With regard to your 'go on calculate it' challenge...please read the answer to Mike...then re read it enough times that you understand it
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Re: Spindrift Calculation??

Brown dog, you have missed the point of just about every post I have made in the last two pages and actually took my words and read them backwards...

As far as overlaying artillery to small armms, well let us know how that works out for you when you compare miles to yards. if that were truly the case sniper school will include a stint at both schools, there has to be reason all that cannon cocker data is left out... distance matters. If it was so important they would have doppler'd everything and then some, twice.

the grooves crutch as you called it, was really a paraphrased analogy to simply say the same things you did, that no two bullets do the same thing when affected by variations to that degree. personally, it's all noise and I can careless.

I believe in Shooter Error more than how much I am affected by Spin Drift... which I contend is small.
 
Re: Spindrift Calculation??

LL,

I'm not going to get into bickering.

Failing to grasp that the ballistics for rifle, cannon and artillery rounds all follow the same ballistic principles and Laws of Physics may be the root of your conceptual 'challenge'
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...and just to nit pick...the grooves analogy is not the same...it applies to the Magnus Effect and the movement of the boundary layer around the projectile.

Have fun
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