One of my other hobbies is flying. One night last year I was messing around with my flight computer and doing calculation for crosswind component, with regard to crosswind component we usually refer to charts but flight computers can also do it and very accurately. I did all these calculations to make this table:
It's a little difficult to get the hang of how to use it initially but it's accurate from my knowledge on the subject.
- The inner most numbers are the clock direction of wind
- The middle numbers are clock position translated into heading direction - this was necessary to convert clock position into heading because the flight computer can't do the calculations using clock position - only heading. It's not detrimental to using the chart, only for calculating the figures for making the chart.
- The outer most numbers Represent wind speed - I used a 10MPH wind for ease of calculating percentages - a 10 MPH headwind has no crosswind componant, 90 degree has a full value being 10 MPH, 10 Oclock 8.7 MPH, 11 Oclock is 5MPH etc. and 10 MPH = 100%, 8.7 = 87%, 5 = 50% etc. You can see the degree away from a 90 degree crosswind is all the same no matter if it's a head or tail wind - for example, 10:00, 2:00, 4:00 and 8:00 - they are all 8.7 MPH or 87% of whatever your particular wind speed is. In aviation head or tail wind does matter for what we're figuring out but with the speed of bullets it's mostly a non-issue.
You can see the example of an 18 MPH wind speed from 10:00 - 10 O Clock is 8.7 MPH or 87% therefor the calculation is 18 MPH X 87% = 15.66 MPH. Now you can get your wind doping into an easier 3 or 9 O Clock figure. If you know your deflection with a 10 MPH wind from 3 or 9 and you're in the field you can make a quicker calculation this way.
If you think it's something useful you could add it in somewhere maybe.
Thoughts?
Wayne
ETA: I forgot the 'proof' that it works and the final figures.
Using the example of 18MPH wind at 10 oclock being the equivelent of a 15.6 MPH 3 or 9 oclock crosswind -
If you know your wind deflection of a particular load at 10MPH at 3 or 9 oclock -
Example load: 75 Gr 6mm @ 3100 FPS, 270 BC, 10 MPH 9 Oclock wind, deflection at 500 yards= 42.2 inches
Using the 15.6 MPH conversion number, you move the decimal over one position making it 1.56 since 15.6MPH = 156% of 10 MPH
Since our 9 O Clock wind deflection of 10 MPH is 42.2" we times it by 1.56 and that equals 65.8 inches
I double checked it on Sierra I6 and changed the same load from a 9 oclock 15.6 MPH wind to a 10 oclock 18 MPH wind and that number was 66"
This system/formula doesn't offer anything that a ballistics program doesn't but it does allow you a way to bring that same accuracy into the field without an electronic device to rely on and once you get the hang of it it's a fast calculation.
Actual wind speed = A
Wind direction value = B
Known load deflection at known distance = E
A X B = C
C X .1 = D
D X E = Actual deflection.
