Rearranging eqn: P+A=R+(A^2+R^2-(2*A*R*cos(theta))^(1/2))

[py97as]

Hi,

I've been trying to rearrange the following to make R the subject and I've not had much luck. P, A, and theta are all known.

P+A = R + (A^2 + R^2 - (2*A*R*cos(theta))^(1/2))

Any help would be much appreciated.

The application is to find the real range of a reflector in a radar environment where R is the range of the reflector, A is the range of the aircraft, P is the path length difference between the direct aircraft signal and the reflected signal and theta is angle between the reflected path and the direct path.

Thanks,

Ade.

 

[Denis]

Re: Rearranging an equation

I've been trying to rearrange the following to make R the subject and I've not had much luck.
P+A = R + (A^2 + R^2 - (2*A*R*cos(theta))^(1/2))

Make it less wieldy: let C = cos(theta) :
P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2)
P + A - R = (A^2 + R^2 - 2*A*R*C)^(1/2)

Square both sides and you'll have no problems...

 

[py97as]

Re: Rearranging an equation

I've been trying to rearrange the following to make R the subject and I've not had much luck.
P+A = R + (A^2 + R^2 - (2*A*R*cos(theta))^(1/2))

Make it less wieldy: let C = cos(theta) :
P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2)
P + A - R = (A^2 + R^2 - 2*A*R*C)^(1/2)

Square both sides and you'll have no problems...

Got it! Nice one, thanks

 

[mmm4444bot]

Make it less wieldy: let C = cos(theta) :

P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2)

Hi Denis:

Did you intend to make the expression A^2 + R^2 - part of the radicand?

~ Mark :?

 

[mmm4444bot]

Got it! ...

Hello Py As:

Whadda ya got?

~ Howard I. Noe :?

 

[Denis]

Re:

Make it less wieldy: let C = cos(theta) :
P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2)

Hi Denis:
Did you intend to make the expression A^2 + R^2 - part of the radicand?
~ Mark :?

Yes; shown is (A^2 + R^2 - (2*A*R*cos(theta))^(1/2))
I assumed that py97 meant that, due to outside brackets not really required;
in other words, I assumed he meant (A^2 + R^2 - (2*A*R*cos(theta)))^(1/2)
(in order for leftmost outside bracket to be required).
However, if he meant A^2 + R^2 not to be part of radicant, then we'd have:
P + A - R - A^2 - R^2 = (- 2*A*R*C)^(1/2)
Due to the mess from squaring left side, I doubt that this was what was meant...
but who knows!

Anyhoo...A^2 + R^2 - 2AR(cos(theta)) looks too much like "law of cosines";
so it sure looks like the whole shot was intended to be part of the radicand.

 

[mmm4444bot]

Yes ...

I assume that the original poster intends what he typed, until proven otherwise.

Cheers,

~ Mark

 

[py97as]

Re: Re:

Make it less wieldy: let C = cos(theta) :
P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2)

Hi Denis:
Did you intend to make the expression A^2 + R^2 - part of the radicand?
~ Mark :?

Yes; shown is (A^2 + R^2 - (2*A*R*cos(theta))^(1/2))
I assumed that py97 meant that, due to outside brackets not really required;
in other words, I assumed he meant (A^2 + R^2 - (2*A*R*cos(theta)))^(1/2)
(in order for leftmost outside bracket to be required).
However, if he meant A^2 + R^2 not to be part of radicant, then we'd have:
P + A - R - A^2 - R^2 = (- 2*A*R*C)^(1/2)
Due to the mess from squaring left side, I doubt that this was what was meant...
but who knows!

Anyhoo...A^2 + R^2 - 2AR(cos(theta)) looks too much like "law of cosines";
so it sure looks like the whole shot was intended to be part of the radicand.

Sorry about that extra bracket guys, P + A = R + (A^2 + R^2 - 2*A*R*C)^(1/2) is what I intended and does as suggested include the cosine law.

The end result being: R = (P^2 +2PA)/(2(P+A-AC))

Thanks again for the help.

 

[mmm4444bot]

Sorry about that extra bracket ...

... Thanks again for the help.

You're welcome, but please understand that the issue is not with the extra parentheses; the issue is with the positioning of parentheses.

You typed (2*A*R*cos(theta))^(1/2). This means the following.

\(\displaystyle \sqrt{2 \cdot A \cdot R \cdot cos(\theta)}\)

Cheers,

~ Mark