Separate names with a comma.
Discussion in 'Rifles, Bullets, Barrels & Ballistics' started by bigsal5353, May 3, 2004.
what is the benefit of each and why are there 2 different twist?
Right-hand twist is best for the northern hemisphere, left-hand for the southern.
Coriolis effect, coupled with gyroscopic precession and the effects of nutation cause the bullet to destabilize and predetonate if the correct twist is not chosen for the hemisphere in which you are shooting.
If you are shooting across the equator, you are simply out of luck...
[ 05-03-2004: Message edited by: STL ]
Is there any literature that documents what you are talking about ? Is this something the average shooter needs to consider or just the 1000yrd bench people ??
You just got your chain yanked...hard...
I'm sorry guys - the question struck me funny, thus the silly answer...
Truth is, I know of no benefits or downside either way, but others may have a much more informed opinion...
Now back to analyzing coriolis-induced northern-hemispheric clockwise toilet swirling...
not to be smarta*s but is that a legit answer? if it is thankyou for answering.. i have no idea and was just wondering. does anyone else ahve a comment
In order to provide some small bit of helpful information - rather than continuing my wanton silliness - here's a link to a bit of history on the subject...
Also, as I recall, the vertical component of wind drift, outlined by Harold Vaughn in his fantastic book "Rifle Accuracy Facts", will function opposite for left-hand twist barrels than for right-hand.
Most shooters are unaware of the vertical component of wind drift, and their shooting technique is not so well-developed as to display it. So, the answer for the vast majority of shooters is that there is no real difference between right and left hand twist rifling.
Spin drift will combine/stack up with coriolis error, same direction ( to the right) I mean, for a right twist for us northern shooters.
You might find this informative:
Originally posted by S1.
<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>S1
Member # 963 posted 12-08-2002 01:44 AM
I want to share a simple formula to estimate yaw of repose. To derive the exact formula for your gun and specific handload, one must collect data in a lab environment that none of us can afford to build. Then you would have to have a serious computer with you in the field to make atmospheric inputs and inputs on the concentricty of the bullet that very few have the ability to measure. My point is it is too impractical for the average physicist much less the average shooter.
So I caution you that this is an approximation simplified from a mid point in bullet b.c., gyroscopic stability factor, and conditions. Yaw of repose is a result of a law of gyroscopic precesssion affecting a bullet spinning 180,000 or more Rpm, being acted on by the force of gravity.
Any gyro (bullet) when unconstrained, will move 90 degrees to the force acting on it. You may prove this by spinning a light plastic top and waiting for it to stabilize in one spot on a table. Once it is stable and spinning on the same spot, gently blow on the top directly away from you. You will observe the top moving away and then going left or right depending on the rotation direction. Gravity is similar to your breath in that it places a force on the gyro without constraining it. Right hand twist barrels cause a rotation that interacts with gravity to produce drift to the shooter's right.
Here is a simple formula that should get an ultra long range set up within 1/4 MOA, provided he is not way over stabilizing the bullet.
[6(Y/200)(Y/200)] divided by 25.4
Y = target's actual distance in yards
your answer or product will be in inches of drift from a 100 yard zero, no wind
These laws of physics are real and you must include them to reliably make hits past 1000 yards.
Follow through matters----- SEE THE FIRE....!!!
Member # 963 posted 12-08-2002 11:39 PM
Lets work from the paranthesis out and see what happens.
1500 divided by 200 equals 7.5. We do this twice and multiply them so:
7.5 times 7.5 equals 56.25 and now we multiply this by 6, and it equals 337.5
Now we divide 337.5 by 25.4 and get our answer:
13.2874 inches of yaw of repose at 1500 yards.
I hope I did not make a typo in the formula, and I apologize for not being more familiar with the word processor software. I have not figured out how to type exponents and other symbols with this web stuff.
Follow through matters----- SEE THE FIRE....!!!
Member # 963 posted 12-14-2002 12:53 PM
To convert 13.2874 inches to MOA at 1500 yds.:
13.2874 divided by 15 = .8858266
Then divide .8858266 by 1.047 = .846 MOA
The coriolis effect strongly corellates to Global position, Azimuth and inclination angle of trajectory, and flight time. It is way complicated and not that significant for flight times under 1 second. I will need more time than I have today to get into this topic. <HR></BLOCKQUOTE>
Mike posted this also:
<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>Mike
Member # 1075 posted 01-19-2003 07:25 PM
In "netspeak" your formula would look like this-
((Y/200)^2)*6) / 25.4
If the nucleus of a sodium atom were the size of a golf ball, the outermost electrons would lie 2 miles away. Atoms, like galaxies, are cathedrals of cavernous space. What feels solid about a tabletop is that the electromagnetic fields set up by atoms in the table repel similar fields in your fist. Matter is energy.
Posts: 1 | From: Texas | Registered: Jan 2003 | IP: Logged <HR></BLOCKQUOTE>
I typed this out for a 750 yard shot to help myself memorize the math.
750 / 200 = 3.75
750 / 200 = 3.75
3.75 x 3.75 = 14.0625
14.0625 x 6 = 84.375
84.375 / 25.4 = 3.321” of drift
3.321 / 7.5 .4429
.4429 / 1.047 = .423 MOA correction at 750 yards needed.
I was actually thinking about this the other day. (I just got McCoy's book)
For northern hemisphere (especially Canada) where the horizontal Coriolis effect can amount to 10-12 inches at long range, it might be worth having left-handed twist which would cause a similar amount of spin drift in the opposite direction. Since both effects increase similarly with distance, they would roughly balance out over a wide range of distances.
I've thought of that too but, I believe coriolis effect has much less effect than does spin drift and would complicate the spin drift calculation, depending on the bearing one was shooting.
Blaine Fields ballistic program, "Precision Shooters Workbench" calculates both coriolis and spin drift.
My thinking was that since the two effects aren't coupled (coriolis is just due to the reference coordinate frame) they can just be treated additively. The ranges I use are facing North so there is no vertical Coriolis effect. I agree that the spin drift is somewhat greater in magnitude, but it might be nice to have a smaller combined correction. Also, the data in McCoys book shows that the horizontal Coriolis effect is essentially independent of the direction of fire. (At least to 1500 yrds)
What I meant by complicate the spin drift calculation was, if you figured one to cancel the other out, but it really doesn't completely cancel spin, nor would it be the same depending on the bearing fired.
The effect of spin drift, in my mind, will always need to be figured seperately, and yes, the figured coriolis correction with a left twist would require you to dial less to compensate for spin, rather than adding to the spin error.
I've never heard of vertical coriolis, that's new to me. Also new to me is the statement that "horizontal" coriolis effect is independant of direction of fire, bearing I'm guessing. That doesn't seem to make any sense?
If this were to be the case, treating them additively as you say does make sense.