Mastering Archery Adjustments: Solving for B with Known Variables A, C, and D

[wvguy8258]

Hi,

It seems I've forgotten more of my algebra than I thought. I am putting together a spreadsheet to help a few archers determine how to change their point of aim for up- and downhill shots. I would like to reduce/rearrange the following expression so that A and B are not found on the same side of the equality, but after 30 minutes can't seem to do it. Ideally, I would like B to be a function of A,C, and D which are known in this setting. Any help is appreciated, even more so if the steps are sketched. -Seth

tan (A) = tan (B) - (9.8*cos(A)*C)/(2*(cos(B)*D)²)

 

[verty]

I couldn't solve for B, I could get it to the form

f(B) = C / g(B)

but not any simpler.

 

[yuiop]

Hi,

It seems I've forgotten more of my algebra than I thought. I am putting together a spreadsheet to help a few archers determine how to change their point of aim for up- and downhill shots. I would like to reduce/rearrange the following expression so that A and B are not found on the same side of the equality, but after 30 minutes can't seem to do it. Ideally, I would like B to be a function of A,C, and D which are known in this setting. Any help is appreciated, even more so if the steps are sketched. -Seth

tan (A) = tan (B) - (9.8*cos(A)*C)/(2*(cos(B)*D)²)

Surprisingly this is a transcendental equation that appears to be solvable, (as long as D/cos(A) is not zero).

The solutions are:

B = (1/2)*(asin(9.8*C*(cos(A))^2/D^2+sin(A))+A)

and

B = (1/2)*(-asin(9.8*C*(cos(A))^2/D^2+sin(A))+A-pi())

You will have to determine which root makes sense in your context.

If that fails use:

C = (tan(B)-tan(A))*2*(cos(B)*D)^2/(9.8*cos(A))

and use the goalseek function of your spreadsheet to find the value of B that gives you the known value of C.

 

[disregardthat]

If you want a go at it yourself, try isolating cos B, while remembering that tan B = sin B / cos B = +-root(1-cos ^2B)/cos B. The sign depends on B, but you will have to take care of that yourself. This will probably end up in a quadratic in cos ^2B, which you can then solve for and after that isolate B. Depending on the domain of B, you might have to consider several cases.

 

[wvguy8258]

Thanks everyone. I feel a little better now that I see solving for B was not extremely trivial. This will be very helpful.

 

[yuiop]

If you want a go at it yourself, try isolating cos B, while remembering that tan B = sin B / cos B = +-root(1-cos ^2B)/cos B. The sign depends on B, but you will have to take care of that yourself. This will probably end up in a quadratic in cos ^2B, which you can then solve for and after that isolate B. Depending on the domain of B, you might have to consider several cases.

I can confirm that the equation does end up as a quadratic in (cos(B))^2. Nice key observation that tan B can be turned into cos B form, that makes the equation solvable!

BTW wvguy, does the equation appear to work all right?