Crosswind lesson learnt

[kiwi3006]

I went hunting on the weekend, saw 7 chamois but all does and young ones so nothing shot.
I ended up taking a shot at a rock at 615 yards. We were at the head of a large valley shooting from one side to the other. The saddle was to our left and the main valley to our right. 10 mph wind right to left.
Came up 10 MOA and right 2 MOA (should have dialed 3 MOA).
Took the shot and hit approx 2 MOA high and 1 MOA left.

I couldn't work out why I was so high. About half an hour later I remembered I should have allowed for a vertical component from the wind.
The slope leading up to the saddle was aprox 26 degrees. This would have given just under 0.5 wind strength which would have lifted the bullet 1.5 MOA which would have accounted for most of the high shot.

Lesson learnt and stored, just as well it was a rock!

Stu.

 

[Michael Eichele]

I am confused.

26 degrees would account for about 2 MOA at 615 yards. You say you were 2 MOA high. Did you compensate for the 26 degree angle?

It seems like the thermals in and of themselves accounted for the high impact rather than the vertical wind component which should have been minimal.

 

[kiwi3006]

Hey Michael, I knew this explanation was going to be confusing. The crosswind was 90 degree right to left at 10 mph AND was coming UP a 26 degree slope. I allowed for the right to left component but forgot to allow for the wind coming UP the slope lifting the bullet.
A 10 mph wind coming UP from under the bullet at 26 degrees should give about 1.5 - 2 MOA lift over 615 yds ( a 90 degree 10 mph wind gives 3 MOA at 615 yds).

Hope this helps to clarify the situation.

Stu.

 

[Michael Eichele]

Hey Michael, I knew this explanation was going to be confusing. The crosswind was 90 degree right to left at 10 mph AND was coming UP a 26 degree slope. I allowed for the right to left component but forgot to allow for the wind coming UP the slope lifting the bullet.
A 10 mph wind coming UP from under the bullet at 26 degrees should give about 1.5 - 2 MOA lift over 615 yds ( a 90 degree 10 mph wind gives 3 MOA at 615 yds).

Hope this helps to clarify the situation.

Stu.

That makes more sense!

Thanks

 

[angus-5024]

can someone point me to a link where I can learn how to calculate for vertical wind deviation, or do I just do it on a ballistics program?

 

[kiwi3006]

The vertical movement is the same as the horizontal movement.
Just pretend the wind is coming from the side calculate how much it would move the bullet and add or subtract from your elevation

Shawn Carlock explains it well in this article Reading The Wind

it is towards the end of the article.

Stu.

 

[blkdog]

I'm an electrical contractor by trade and this past weekend I had an eye opener as to reading wind. I was called out to consult on some small wind generators in SW idaho. These were placed on a high knob between two opposing ridges. The towers were set in a triangle pattern on the top,but at the edge of the knob.As we looked at the installation, I began to watch the direction the blades were pointing. Although the towers were only 150 feet apart,at times the blades would be as much 20 degrees apart! Now try to figure how to shoot across this ridge or up or down the slope! The wind direction was so unpredictable that it made me think that sometimes it is just luck that makes a long range hit happen, I know that they do on a semi regular basis but after watching that wind I had to give my wind reading skills another look.