Wind Drift formula?………Again.

[Bushmaster55]

I know this subject has been hashed over on this and I’m sure many shooting sites across the web. I don’t consider my self a novice shooter and by no means an expert. I have been on the gun since my younger years and have comp. experience at Camp Perry several years in a row with everything from a 30 carbine, my Granddads 1903-A3, M1, M1A, and an AR. I’m an avid white tail and predator hunter. Not patting myself on the back just trying to give you an idea were most of my experience lies. I kinda blame (Thank) all you for getting me into this LR stuff, although my wife feelings probabley differ.
OK enough of the BS. My question stems from a thread on SH about wind and without naming names an equation/formula was posted to calculate wind drift. I am familiar with this formula from reading of it on this site but with multiple constants used for varying yardages The formula as posted (Distance divided by 100 times wind velocity at mid range (in mph) divided by constant equals favor.) Further in this thread and the particular example the constant of 10 was used to estimate favor. My question is what is correct? Does the constant change with distance or stay the same (10) as in this example?
The previous formula I have seen uses different constants per yardage.

100 – 500 yd. = 15 constant
600 yd = 14
700 yd = 13
800 yd = 13
900 yd = 12
1000yd = 11

Sample
Distance to target 500 yd.
Wind 8mph @ 3 o clock

Formula for a 500 yd. target ( 500 / 100 x 8 / 15 = 2.66 inch right wind correction)
Formula for a 900 yd. target (900 /100 x 8 /12 = 6 inch right wind correction)

I’m well aware of the many variables that affect bullet trajectory
I’m of the K.I.S.S. crowd and while I find the scientific explanations to some of this stuff interesting ( not really) I’m looking for simple and fast. I’m not going to die if I don’t put one thru that coyote at 650 yd. but it sure will put a smile on my face. Thanks

 

[Sterling Shooter]

Re: Wind Drift formula?………Again.

You appear to be citing the formula I had mentioned. Constants do change for distance, that's to say, the constant is not constant. The constant is also not likely to be exactly 10; but, for government work, it works relatively well for any distance to about a 1000 yards, when shooting high B.C. bullets at high velocity. Using the number 10 when perhaps 11 or 9 might suit better permits math in a heartbeat while "on-the-gun". On-the-gun is important since figuring a favor quickly is as paramount to a good hit as is favoring for wind correctly, beginning at distances around 300 yards. Now, here's the rub, if you've got a mil-dot reticle you would use a constant of something like 33 to 37 to get an answer in mils of favor. Trying to divide by these numbers on the gun is not as fast as using the constant of 10, which gets you an answer in MOA. Of course even if your reticle is graduated in mils, if your windage adjustment is graduated in MOA you can click rather than favor, but this is generally slower and invites error.

 

[Bushmaster55]

Re: Wind Drift formula?………Again.

Sterling I was hoping you would find this question as you are correct it was your remarks that I was commenting on. I knew I should have included more info. I’m working with a 308 -168 & 175 SMK’s at around 2650/2700 fps and may fool with Hornadys A max’s which have a little better BC. I completely understand the theory between simple math and a fast favor. Like I explained no ones shooting back in my case (ground hogs ,coyotes) and when in place I can estimate wind vel, direction and such for the location but not the target since it may appear anywhere. So fast in some cases is necessary. Since you all got me all jacked up on this long range thing and I have been able to work more on long range shots. Coyotes often present themselves (if I set up right) at 400 to 500 yds. and simply sit down and look for what’s making all the commotion. This does give me some time to calc. Wind as in NW Indiana this time of year is most every day. I believe wind is not an exact science though some seem to try to explain it as such. I see to many variables for my application to go scientific with my calcs. SS I like your approach and your comments in that particular thread and expect your next comment to me to be “you just need to go shoot in the wind” which is what I need to do. There’s nothing that can replace practice and experience.
Sterling, If I get this right what your explaining is constants do change with distance but for the sake of time using 10 for the constant should produce the correct answer in an acceptable time frame to get the shot off before the conditions change again. Not trying to put you in a trick bag and believe me I’m sure you have much greater knowledge and experience in this than I.

 

[kraigWY]

Re: Wind Drift formula?………Again.

I'm a simple man, not much of a math wizzard, I can't do division in my head that fast, nor do I want to take a slide rule to the firing line.

I have been shooting High Power since 1977, during that time I have never seen a "steady wind". Watch a range flag. They do not get a single "angle" and stay there. They are constantly moving from X degrees to X +/-. They never stay put.

I know that, looking at balistic programs, 14 constant is probably more accurate.............IF.. the range flag held still.

I can't hold an the X ring at 600 with iron sights, meaning I can't hold the difference between a constant of 14 and 10.

I do know the faster I shoot, the better my scores because the faster I shoot, the less the wind changes. It's not going to hold still while I do the math on a calculator.

I have to keep it simple. Even my math challenged mind can figure that dividing any number by 10 is more simple, and faster then dividing a number by 14.

I may be a tad off the 14 constant numbers, but the time it takes to divide R*V/14 as opposed to R*V/10, the wind is going to change more then the difference between the two constants.

Besides, I use half min clicks on my service rifles. 1/2 min clicks don't allow for the difference between costants of 14 and 10.

If you can do the math in your head, by all means use 14, I can't, I have to keep it simple.

 

[Lindy]

Re: Wind Drift formula?………Again.

I have a version of that formula which gives wind holds in mils - which is what my scope reticle is calibrated in. You can derive your own by multiplying those constants by 3.44.

The point is, the formula is only a place to <span style="font-weight: bold">start</span>, as it assumes that the wind is exactly 90 degrees to the line of fire, that the wind is constant along the entire path of the bullet, and that you know what the wind speed is over that distance to within 1 mile per hour.

All of those assumptions are probably wrong. On shots beyond 700 yards, you can pretty much bet they are egregiously wrong.

My advice to the OP is don't get wrapped around the axle trying to get it perfect - because you can't. Use whatever formula you want to make the first shot - and correct from there.

 

[Sterling Shooter]

Re: Wind Drift formula?………Again.

The wind formula as stated, whether divided by a constant to yield an answer in MOA or mils, works no matter what the direction the wind is blowing. Using the wind clock, figure wind as having a full, half, or no value. For example, a 10 mph wind from 11 to 5 would be described as a half value wind and 5 mph would be factored into the formula.

 

[Sterling Shooter]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bushmaster55</div><div class="ubbcode-body">
Sterling, If I get this right what your explaining is constants do change with distance but for the sake of time using 10 for the constant should produce the correct answer in an acceptable time frame to get the shot off before the conditions change again. Not trying to put you in a trick bag and believe me I&#146;m sure you have much greater knowledge and experience in this than I.
</div></div>

You generally do what you have time to do. Lindy's mentioned approach is accurate, and what Kraig alluded to is no doubt fast. For me, shooting as issued M16 A4 irons, graduated in MOA, I do it in a manner that I perceive complements the requirement. If it's speed, I'll favor, converting MOA to inches. If accuracy is paramount, I'll click. With favors, I can stay on the gun. When shooting in slow fire HP competition where adjusting the sight for windage will assure better accuracy, I'll come off the gun and click. Since clicking can invite error, keeping up with it is a must, meaning notations in the data/score book.

 

[Lindy]

Re: Wind Drift formula?………Again.

What Sterling Shooter is saying is basically what I call Supernaw's Law - named after Irwin Supernaw, the father of country singer Doug Supernaw: "The only correct answer to any nontrivial question is, 'It depends.'"

 

[Bushmaster55]

Re: Wind Drift formula?………Again.

I appreciate all the responses and I’m gonna play with the formula and different constants and figure out what works for me and Kraig I hope # 10 works best cause I’m with you on keeping the math simple especially in a hunting application. And SS I already figured out the part about entering the velocity after you figure value of it, only make sense. Thanks again.

 

[kraigWY]

Re: Wind Drift formula?………Again.

DFOOSKING

I never heard that before, (and I've been shooting a long time).

I just checked you theory with a Ballistic Chart using 168 SMKs at 2550 fps. That's pretty damn close to 600 anyway.........

And fast, I like fast.

 

[jasonahull]

Re: Wind Drift formula?………Again.

Man watching the new dvd on precision rifle, I got seriously lost in his math. It took a couple of times for it to make any sense. If my ballistic computer ever quits I am in trouble.

 

[Sterling Shooter]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: a-hull</div><div class="ubbcode-body">Man watching the new dvd on precision rifle, I got seriously lost in his math. It took a couple of times for it to make any sense. If my ballistic computer ever quits I am in trouble.</div></div>

I have empathy for you. Leaning how to shoot, without a highly qualified mentor who can assist with shooter/target analysis, makes learning, from on-line materials, DVD's, and books on marksmanship, somewhat sketchy. But, here's one thing for sure, if hitting a target requires a calculator you may want to re-think it all. After all, if you're out there by yourself and are most certainly going to engage an unknown distance target in conditions out of your control, good hits will likely require you can properly adjust the sight and favor from calculations you can do in your head. Study of external balistics with emphasis on the effect of gravity, drag, wind, weather, and slope will get you off to a good start.

 

[Lowlight]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: a-hull</div><div class="ubbcode-body">Man watching the new dvd on precision rifle, I got seriously lost in his math. It took a couple of times for it to make any sense. If my ballistic computer ever quits I am in trouble.</div></div>

That is because it requires a shifting scale to even come close. He sells these concepts as if they are unique as well as universal which they are not. in many cases they are just elements from the software, I would take the information presented as something that works for one guy and few others.

 

[LoneWolfUSMC]

Re: Wind Drift formula?………Again.

Something that is lost on some shooters is there is no "x+y=Z" answer to wind.

Wind is a constantly changing variable. At some point experience and "feel" takes over where math left off. You need to pay attention to how it feels.

The math is important, but tuning your senses to the wind changes is equally so.

 

[Pointman308]

Re: Wind Drift formula?………Again.

I know how to calculate for wind using moa. How do you do it for mils? Just multiply 3.48 x moa for the mil solution or is there a strealine way of doing it for mil adjusted scopes? Perhaps there is a different constant for mils?

 

[jasonahull]

Re: Wind Drift formula?………Again.

Agreed. I really enjoy shooting. Hate math. I dont want to spend as time on it as he has. Those dudes in the video had all the math and still missed. Having the math figured and reading the wind correctly are completely different to me.
I cant imagine my self looking through my scope and trying to figure all of that out to make a shot. It would take an awful lot of practice.

 

[RemingtonSPS]

Re: Wind Drift formula?………Again.

A few F-Class matches with a standard load did more for my wind skills than anything else. Doing a few long range tac matches with a spotter helps too. Alot of it is experience.

His wind formula made alot of sense to me. What he was attempting to do was make it simple enough to do in your head. I found a way to make it work with MOA instead of MIL and i want to try it next time shooting just for giggles to see how close it is to my gut feel. His formula is .1 mils for every hundred meters in a 4mph wind (with .308win/175SMK), essentially moving the decimal for every hundred meters. For example the .308 win. At 500 meters in a 4mph wind is a .5 mil hold, .6 mils at 600, etc. If it was an 8mph wind you would double it. And at 700 meters and out you add a tenth, I.E. 700 meters 4mph wind equals.7 mil+.1 mil for a .8 mil hold. 800 meters would be a .9 mil hold, etc.

It doesn't have to be a .308 win or even MILs either. For MOA you can find what wind speed gives you a five quarters (5/4) hold at 500yds for YOUR bullet using your PDA. For my load it was 6mph. So at 500yds i would hold 5/4moa (1.25moa) for every 6mph wind value. At 600yds it would be 6/4 (1.5moa) etc. And at 700yds and out you would do 7/4 (1.75moa) and add .25moa for that distance and every hundred yards past that, just like he added .1mil onto his formula for 700 meters to give him a .8mil hold. For every 6mph wind at 700yds the wind equals 7/4moa plus 1/4moa (because at 700yds and farther you add .25) which gives me 8/4 or 2moa wind hold for every 6mph with MY bullet. I think its easy, i can do it in my head or off the fly without consulting a wind chart or PDA. Wether or not it will work in the field i will find out soon enough. They didn't seem super consistent with their hits so im a little skeptical.


He called it 3rd grade math so i don't think it was supposed to be complicated

 

[Lowlight]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: MuleHunter</div><div class="ubbcode-body">A few F-Class matches with a standard load did more for my wind skills than anything else. Doing a few long range tac matches with a spotter helps too. Alot of it is experience.

His wind formula made alot of sense to me. What he was attempting to do was make it simple enough to do in your head. I found a way to make it work with MOA instead of MIL and i want to try it next time shooting just for giggles to see how close it is to my gut feel. His formula is .1 mils for every hundred meters in a 4mph wind (with .308win/175SMK), essentially moving the decimal for every hundred meters. For example the .308 win. At 500 meters in a 4mph wind is a .5 mil hold, .6 mils at 600, etc. If it was an 8mph wind you would double it. And at 700 meters and out you add a tenth, I.E. 700 meters 4mph wind equals.7 mil+.1 mil for a .8 mil hold. 800 meters would be a .9 mil hold, etc.

It doesn't have to be a .308 win or even MILs either. For MOA you can find what wind speed gives you a five quarters (5/4) hold at 500yds for YOUR bullet using your PDA. For my load it was 6mph. So at 500yds i would hold 5/4moa (1.25moa) for every 6mph wind value. At 600yds it would be 6/4 (1.5moa) etc. And at 700yds and out you would do 7/4 (1.75moa) and add .25moa for that distance and every hundred yards past that, just like he added .1mil onto his formula for 700 meters to give him a .8mil hold. For every 6mph wind at 700yds the wind equals 7/4moa plus 1/4moa (because at 700yds and farther you add .25) which gives me 8/4 or 2moa wind hold for every 6mph with MY bullet. I think its easy, i can do it in my head or off the fly without consulting a wind chart or PDA. Wether or not it will work in the field i will find out soon enough. They didn't seem super consistent with their hits so im a little skeptical.


He called it 3rd grade math so i don't think it was supposed to be complicated

</div></div>

Rule of 9s comes to mind and the British Method of wind calls, its all just a take on those methods. It's not new.

<span style="font-weight: bold">British formula</span>
10MPH wind is your base wind. (wind speed that equals 1moa per 100 yd)
Off of base wind 2-3 = light 5 = medium 10 = base 20 = heavy

Example
Range 600, velocity 10mph = 6 moa
Range 600, velocity 5 mph = 3 moa
Range 600, velocity 2-3mph=1.5 moa
Range 600, velocity 20 mph = 12 moa (and call for indirect fire)

<span style="font-weight: bold">Rule of Nines </span>
Rule of nines is nothing more than using 9 as the quarter mil factor. So if yo multiply range in hundreds time wind speed without dividing by ten ou get a 35 in the above example. Since 9 = .25 mils, 18 = .5 mils, 27 = .75 mils and 36 = 1 mil, then 35 equals 1 mil same answer.

Next is a direct to mil conversion based on BC.

The BC for the standard .308 is .4xx so you use a 4 mph window.

100 = .1 mil at 4 mph
200 = .2 mil
300 = .3 mil
400 = .4 mil
500 = .5 mil
600 = .7 mil due to velcity bleed off and correcting the estmate
700 = .8 mil
800 = .9 mil
900 = 1 mil
100 - 1.1 mil

Every four mph window then becomes a multiple so 16 mph wind at 500 is 4 times .5 mil or 2 mils.

Thsi can be used for 5.56 with a BC of .3xx, using 3 mph windows, the 300 Win Mag with .5xx BC using 5 mph windows and so on. Adding .1 mil to the window for each mph, ie 5 mph at 500 is .6 mil will get you on the target.

When you don't know where to look anything present to you for the first time appears new and unique and potentially effective

 

[RemingtonSPS]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Lowlight</div><div class="ubbcode-body">When you don't know where to look anything present to you for the first time appears new and unique and potentially effective </div></div>

Agreed. I haven't had any formal training. I read than try to put it into practice to test the results. I tend to favor experience. Too much 'doctrine' without practice can jam a guy up.

 

[jasonahull]

Re: Wind Drift formula?………Again.

Yeah and when guys use words like rules and formulas, they make it seem like you should expect to get first round hits, or atleast I read that in to it, so the getting confused by all the math does get in side my head. Like I said, I hope I never lose my ballistic program. Or the printed chart. I havent read this whole thread, but am I the only one who thought the wiz wheel wasnt so simple?

 

[RemingtonSPS]

Re: Wind Drift formula?………Again.

If you take the ballistic programs dope literal for the wind your gonna have problems too. If you can get a reference from a PDA you can get a reference from your head.

 

[Augustus]

Re: Wind Drift formula?………Again.

Here is something i use in a pinch on a 308, 175 gr SMK at 2650. Wind speed in MPH times Distance in yds then double it answer is in mils. 700 yds in 7 mph wind. Move decimal three places to left on distance .7 times 7 = .49 times 2 = .98 mils Atrag says 1.13. This will usually get you to .1 or so. Not perfect but close. Once you do this afew times you can get prety quick at it.

 

[Augustus]

Re: Wind Drift formula?………Again.

after a little bit you learn not to worry about the decimals just deal with the numbers, when you get into wind speeds 10 and above the formula I gave gets off by one decimal place, so you learn to ignore it. Also you can either round off to the nearest 100 yds on the distance or you can figure the even hundreds above and below the actual distance and split the difference. This may sound confusing but its not. If I can do it anybody can. If you practice you will get to the point its almost automatic. Single digit wind is easier to do. Also when the wind gets above twelve mph and the distance gets 700+ you have to add 1 or 2 tenths. 850 or above stick on .4

 

[Augustus]

Re: Wind Drift formula?………Again.

Oh, for drops with a mil-dot scope, same load mentioned above. Distance in yds - 2. Example: 450 yds - 2 = 2.50 mils
Example: 6oo yds - 2 = 4.00 mils

This works very well out to 600 yds at 2650. Again Ive learned to add a couple of tenths where needed past 600. These methods are not exact but you can carry this around in your head and you dont have to worry about the batteries going down. I know I took liberties with the decimal, dont worry about it, accept it and move on.

 

[Lowlight]

Re: Wind Drift formula?………Again.

Have to say, I have metric ton of time using a 175gr bullet at 2650fps and none of your numbers are right. Don't know if they are examples but 4 mils at 600 yards, no way, in the crazy winds of South Texas at a 1000 yards we used 3.5 mils or there abouts, at 400 yards it's closer to .5 mils to .75, 500 was 1 mil, 600 was 1.25 mils, etc.

Not sure if your numbers are just put out there but they're not right for a 308 at all. Not in mils anyway. 600 yards being 4 MOA is closer to correct then mils.

 

[Augustus]

Re: Wind Drift formula?………Again.


Lowlight, are you drunk or something. Go back and carefully read my post. This time try to focus on the content of the posts and not so much about closing in for the kill. There are two distinct subjects in those posts, one adresses a method to use if nothing else is available to figure winddrift on a 308, that was the first post, in the later post I talked about how to figure holdover if nothing else is available. The 4 mils you are so upset about is the HOLDOVER in mils for a 308 at 600yds, it is not (NOT the holdoff for winddrift). As stated this just gets you close, the actual holdover for a 308 at 600 yds is about 4.3 Mils. Past 600 I always add a couple of tenths which would put me at 4.2 mils. Im 0ne tenth of a mil off or 2.16 inches.

 

[Augustus]

Re: Wind Drift formula?………Again.


For 338 Lapua, 338 Edge etc 300SMK 2850 fps. Wind times Distance. Example: 8 MPH wind - 400 yds
8 times .4 = .32 mils

You can simply multiply the distance and wind and get a holdoff for WINDAGE. This works pretty well out to 400, from 400 to 800 add .1 (one tenth). 800 to 1000 add .2 (two tenths). 1000 to 1200 add .3 (three tenths)

These figures are based on 100 yd zero, also they are approximate and as everyone knows the usual fluctuations due to atmospheric conditions apply. If you have good dope use it, if you have a ballistic computer use it. If you get caught without anything else this will work pretty good. I also have the same kind of game to play on the HOLDOVERS for the Edge and Lapua but it is a little more involved tahn the 308 Holdovers.

 

[Pointman308]

Re: Wind Drift formula?………Again.

Your .308 formula seems to be pretty close. And thats a lot easier than moa to me. Thanks for sharing that Augustus. Where did you come up with it?

 

[Augustus]

Re: Wind Drift formula?………Again.

Tinkering with the Atrag on my Nomad. Just enter your info, click on the range card and scroll down through the ranges and you will start to pick up trends. The reason it works so well for the 308 and Edge, Lapua etc. is that they have a parabolic trajectory through a lot of the commonly used ranges. Rds like the fast 22s and 300 Ultra mag a harder to do for the drop because the constant is not an even number and it is harder to deal with in your head. For drops on the Edge and Lapua with the 300 gr. at 2850 from 600 to 1100 yds it is a simple matter of Range - 3. Example: 825 yds - 3 = 5.25 mils, after 1100 the trajectory starts to drop out of the rythem and you have to start adding tenths for each 100 yds or whatever way you decide to band it.

 

[Pointman308]

Re: Wind Drift formula?………Again.

You must be a math wiz in which I am not. If I switch over to mil adjustments I will definity have your wind formula in my log book. Thanks a lot.

 

[Augustus]

Re: Wind Drift formula?………Again.

No Im not a math whiz, as a matter of fact Im just a dumb-ass hillbilly. Get a mil based reticle such as the leupold TMR or the Horus H-25. I use the H-25 with my 338 Edge. Dont worry about the knobs, use the reticle.

Im not talking about match shooting, its for quick hits on 12 by 12 targets.

 

[Lowlight]

Re: Wind Drift formula?………Again.

Drunk No,

Saw it on my iPad and one post blurred into the other, then going off topic when you posted 3 times in a row, it appeared to follow the same course. It never registered the holdover part especially because it doesn't say anything about being from a 100 yard zero, so it really just blurs back to wind because you say, "using the above". It followed too closely to the wind information without a proper separation to register.

Learn how to communicate effectively, it might help get your point across better. Just because you say your a dumb hillbilly, doesn't mean you can't take a lesson on sentence structure.

 

[Pointman308]

Re: Wind Drift formula?………Again.

Thats kind of one of the arguments I have as it pertains to mil based scope adjusments, if what you grew up on was moa based I dont really think they are an advantage. I just talked to a 1st Seargent whos in charge of snipers and is a sniper himself and he made the comment that the Army is going to start using hold overs more and more. Who knows since the Army is such a large machine its hard to tell, but the prospect is interesting. I love the TMR reticle, but the Horus is a little too busy for my likes. Does it obstruct your field of view a little?

 

[Lowlight]

Re: Wind Drift formula?………Again.

Yes the Horus is for the lowest common denominator and most who use it realize within 6 months they can do more with less. In elevated places where you can see the impacts on the ground more clearly if works ok, however obstruct the background and chances are you'll miss the impact behind the reticle or not see it at all.

The horus was designed to help the owner hunt when he was too busy to practice.

Mils are just numbers, same as MOA, think Pennies and Dimes... the reticle is broken up to 10 pennies with a Dime being the Mil Dot and moving out from center that way. It's very simple to use mils. You just read 10 pennies and 1 dime ...

Its good to teach a person mils on the horus, but then its better to move them away to something that offers a better FOV.

 

[Augustus]

Re: Wind Drift formula?………Again.

My God Lowlight, talk about the pot calling the kettle black. Look at the first sentence in your last post. Talk about poor communication. ( the most who it ) Why dont you just go ahead and tell me you hate my guts and anything I have anything to do with.

Pointman, I assume you directed the question about the Horus reticle to me. I like the Horus, I also like the TMR and in a pinch I can use a plain mil-dot to smack things at varied distances and wind calls very quickly and efficiently w/o the use of notes, knobs, or computers. It boils down to a matter of preference. Warning: there may be examples of poor spelling, grammer, or sentence structure lurking in my posts just waiting to cause someone to make a public idiot of himself.

 

[Christopher Rance]

Re: Wind Drift formula?………Again.

Good point on the Horus.

As for wind I use the range-1 or BC relation formula. Both work for me.

 

[Pointman308]

Re: Wind Drift formula?………Again.

I always kind of considered holdovers to be kind of like Kentucky windage in a way, but after saying that I have made some pretty solid hits at 700 to 1000 yards using it. I think that holdovers have a place as far as expediency is concerned, but if we are talking one shot type situations I dial for wind and range and hold with my reticle for movers. Just how I do it.

 

[Bushmaster55]

Re: Wind Drift formula?………Again.

You know guys in some ways this thread went every which way other than K.I.S.S. like I hoped and thought I ask for. I know the question I ask is not to be answered simply. I simply wanted to know if the example that was given with using 10 for a constant was a solid place to start my own experimenting or if the example using different constants for different yardages was a better place to start. I understand there are many so called formulas out there that may work for others and they obviously use. I thank those who suggested what works for them as some of the ideas are very interesting. I also understand many use battery failing electronics to arrive at wind calls and hold overs in MOAs and MILs . And if that’s what they want to use that’s OK with me it’s just “Not for me”.
After I posted this question I received the FDAC (Field Density Altitude Compensator) which I recently found on this site and which has many favorable reviews. I will have to say this is K.I.S.S. at its best if it works as intended. I’ve wasted way more money on other accessories if it doesn’t.
If I could officially end this thread I would cause I feel it has taken a slightly disrespectable air about it when I see a relatively new user calling out someone who’s been around for a decade with thousands of post of which I’ve read many and believe he know what’s he’s talking about. This is just my two cents and you can call me out if you like but I’m not going there, so take your free shot if you want.

 

[Deleted member 10043]

Re: Wind Drift formula?………Again.

<div class="ubbcode-block"><div class="ubbcode-header">Quote:</div><div class="ubbcode-body">100 – 500 yd. = 15 constant
600 yd = 14
700 yd = 13
800 yd = 13
900 yd = 12
1000yd = 11</div></div>

What is known today as the Old USMC Formula for MOA adjustments. Applies to .30 caliber and seems to work with a little practice and it is fun trying it out. It is a battlefield method dating back to the 1903 Springfield. The math constants are for a 10 mph crosswind. 10 mph winds from other directions reduce the wind speed variable by a certain percentage or something like that. Anyway, there is a lot of stuff on the net about it.