This is my actual wind clock system that I use. It's worked for me nearly spot on when I use it. If your results differ, then use your system. I'm just relaying what has worked for me. I have taken shots at the 7:30 wind direction at 1597 yards, and missed due to elevation and not the wind. That was the pig shot I took. I went right over the back of the pig. Wind call was spot on. I'm showing what has worked for me, and certainly will work for other shooters.

Ok, so I’m not nuts after all! Chad, can you tell us who designed it? Not trying to be argumentative, just trying to learn.

Vectors and simple trigonometry say the clock is right.

Please show us those equations, papa. Thanks!

Watching bullets go down range a whole lot doesn't make that clock work in my head. I use the clock, but with different values. And there is a bit of difference in left to right vesus right to left crosses. One of my rifles spin drifts right like no other, and I am not over twisted with it. Just gotta know it does it.

Hard to show without drawing a picture, but here goes. A vector is a force, such as wind, of some magnitude in some direction. Working in two dimensions only, that vector can be described as having two right-angle components (like vertical and horizontal on a graph.) Now imagine yourself sitting in the middle of Chad's clock, with 12:00 straight up. That wind vector coming in at 1:00 is 30 degrees from 12:00. Now lay a 30-60-90 triangle on the clock with the short side horizontal at the top and the longest side at 1:00. If the value of the wind vector on the long side is 2, the short side value at the top is 1, and the side going up the middle is 1.732. When considering wind drift, only the horizontal (90 degrees sideways) component is significant. If the wind vector is 2 and the horizontal vector is 1, then the horizontal component is 1/2 (0.50) of the actual wind vector.

You can do the same thing for each hour position using the same 30-60-90 triangle.

Hmmm, now even more confused.... isn’t a 30-60-90 degree triangle also called a right angle triangle?
If so, according to our old friend Pathagorius, the long side would be 3.99636, making the horizontal component 1/4 or 25%.

Pythagorean Theorem: the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(2x2) = (1x1) + (1.732x1.732) or 4=1+3

Oops, you’re right on that papa, somewhere I was off by 1. I’m still not sure I understand it all, especially since I found the same clock on another site and on the same page, down lower, it showed an entirely different result. To me, logic dictates that each hour; 1:00, 2:00 and 3:00 should be (in order) 33.33%, 66.66% and 100% of full value. Good thing I ain’t makin’ a living at this.

First simple vectors and then basics in forces as a vector.

Technically speaking, a vector is a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another. Favorite example is speed vs velocity: speed is magnitude only and velocity is a vector. Speed is 35mph. Velocity is 35mph, heading north on I45. Length is magnitude only and distance is a vector.
So mathematically vectors can have negative directions and even then that’s a conversation around reference frames. Keeping it simple, in a 2d plane vector analysis is simple trigonometry. So if you have a constant value of a vector (V) then depending on where it ends as pivots arounds the center of the circle, then the length along the x and y axis can be broken up in to COMPONENTS using simple sin and cos functions. The is the Pythagorean theorem that was being referred to earlier. The circle is drawing to geometrically show all cases will have the value of V be the same. So as long as you don’t change the value of V, the X and Y will always be the same for that given degree. Mathmatically you can break up any vector into its components and it only has to do with your reference frame as to what you call X or Y.

So with V,
X = V * sin alpha;
Y = V * cos alpha;
And therefore, tan = X/Y. to math nerd yall some more.

The 30 degree and 45 degree examples were drawn Autocad the I took it off the measurement. In the 30 degree example
433 = 500 * sin (30 degrees), which also sin (30 degrees) = 433/500 = .866
250 = 500 * cos (30 degrees), which also sin (30 degrees) = 250/500 = .500
Btw, if you’re checking this and it doesn’t add up with the trig function then you might wanna check the units angular measurement to mae sure it’s not on radians. If so, you can take the degree value times Pi and divide by 180.





With vectors and how to break them into components are now covered…

All forces are considered vectors. And technically speaking, the wind direction is vector and therefore the drag force exerted on the bullets is in the exact same direction.

So relating to the wind clock is easiest to first view everything in the context of 1.0, or sometime I’ve heard to as the FULL VALUE wind. This actual value is not important now. So whatever the wind does at whatever speed, let’s say you need exactly that speed of wind blowing left to right constantly the whole time across the whole flight time of the bullet then it will move exactly 1.0 mil (or moa but angle so distance doesn’t matter) to the right.
Now let’s say you have a constant 2 o’clock wind that causes. An now, take the math formulas above for the 30 degree example. So if all of the following are true about wind forces (which can reasonably be assumed as true)
- 12 o’clock is in the direction of shot
- 3 o’clock is mathematically called the starting 0 degree point (meaning at 3 o’clock wind direction then alpha; = 0 degrees) as works your way around in the counter-clockwise direction you increase the value of alpha;up to 360 degrees. So the following (something) o’clock = alpha; degrees is 3 = 0 degrees, 2 = 30 degrees, 1 = 60 degrees, 12 = 90 degrees, 11 = 120 degrees, 10 = 150, 9 = 180 degrees, etc.
- You assume that component of the wind in parallel to the shot, ie the Y component does not affect the wind drift. This is just true axiomatically to vectors. Therefore you’re only concerned about the X value.
- Since the sin function oscillates from 1.0 to -1.0, the above reference frame means positive X is to the right and negative values just means to the left
The only thing that remains is that you can do this analysis in discrete intervals, such as the 30 degree separations in a clock. After doing, the space in between are sectioned off basically saying that everything within +/- 15 degrees (or however much you chose) of each interval so the center analyzed values is the one taken.
So now, relating a constant wind speed to a FULL VALUE mil/moa hold and then know the “o’clock” direction will tell you how much of the wind is acting by multiplying the given value in the table by the FULL VALUE hold.