Hornady's ballistic calculator ... something's not quite right!!

[MCMXI]

I've been using Hornady's ballistic calculator located here ...

http://www.hornady.com/ballistics/ballistics_calculator.php

... to work up some load data for the 208gr A-MAX at 2850fps. However, there seems to be a problem with their windage
calculations in terms of the wind drift corrections in MOA at 5mph and 15mph.

My calculations agree with the 10mph crosswind data but I'm not even close on the 5mph and 15mph values. Am I missing
something here or is there a bug in Hornady's ballistics calculator?

Thanks.

 

[TimRB]

The PCB freeware ballistics program agrees with Hornady, so my guess is you're missing something. Just on inspection, drift seems to be increasing linearly with the Hornady program (out to 600, at least) and that seems right to me.

Tim

 

[NCsmitty]

Well roughly figuring at 1000yds and 5mph you got 30" of drift. At 10mph you got 60" of drift and at 15mph you have about 90" of drift. That appears pretty static to me.
Where are you having trouble with the numbers?

NCsmitty

 

[MCMXI]

Tim and NCsmitty, the wind drift values may be fine, but the MOA windage calculations (adjustments) based on the drift are what I'm having a problem with. I'm using the following equation ...

MOA (windage) = [drift (in.)*100 (yards)]/[range (yards)*1.047 (in.)]

Their numbers (for 5mph and 15mph) may be off by a factor of 2.

 

[TimRB]

Ah... I see what you mean now. PCB confirms the drift (inches) and calculates the proper MOA (as you've done, too) but we disagree about Hornady's MOA. I don't quite agree about the factor of two difference, though. Anyway, maybe Hornady does some sort of fancy nonlinear correction for windage.

Edit: No, that can't be it, because the drift in inches works out...

Tim

 

[NCsmitty]

The 5mph moa drift is indeed off by near 2 factor on the negative side and the 15mph moa is off approx. 1.5 to the positive. All rough calculations.

NCsmitty

 

[DaveBeal]

Hornady's drift in MOA is proportional to the square of the wind speed. At any distance, the drift in MOA for 10 and 15 mph are 4x and 9x that for 5 mph. This would make sense if the sideways force exerted on the bullet by the wind was proportional to the square of the wind speed. That might be true, because for a subsonic airplane, aerodynamic drag is proportional to the square of airspeed. But I've never been sure whether that applied to bullets, because of the differences in size and speed.

 

[MCMXI]

Hornady's drift in MOA is proportional to the square of the wind speed.

But Dave, that may be true when calculating the wind drift in inches, but converting inches of wind drift to MOA has nothing to do with the square of the wind speed.

My first thought was that Hornady knows something that I don't and has made an adjustment but now I just think it's an error in their calculator which is odd since I agree with their numbers for the 10mph cross wind.

NCsmitty, "All rough calculations." ... that's what I did ... but I just looked at a couple of the 5mph MOA values and guessed at the factor of 2 error. Very confusing either way since the error isn't consistent.

 

[DaveBeal]

But Dave, that may be true when calculating the wind drift in inches, but converting inches of wind drift to MOA has nothing to do with the square of the wind speed.

I agree with your statement, 1858. I'm not saying that the conversion of inches to MOA involves the square of wind speed. I'm suggesting that, at any given range, drift measured in inches and MOA may both go as the square of wind speed.

 

[DaveBeal]

Now I'm confused. The first equation on this webpage seems to say that the drift distance is proportional to wind speed, not to its square. So maybe the Hornady calculator (and my theory) are wrong.

 

[MCMXI]

Dave, that's an interesting link you posted. It looks like Hornady might be using that first equation (or similar) from Sierra to calculate the wind drift in inches. I'm not able to accurately calculate the bullet time of flight so I just divided the distance by the average velocity to get time of flight. Using that value, I calculated wind drift and it's close to the values listed. However, that still doesn't account for the "incorrect" conversion from wind drift (in inches) to MOA.

Has there been a shift away from the traditional squares rule to calculate wind drift? As you mentioned, wind drift is a linear function of wind speed according to that first equation.

This will have to remain a mystery until I contact Hornady.