TRIGONOMETRY equation derivation [HELP]

[Rihanna]

The signal length from satellite to the earth station (AC) can be found as


2(H)/[{sin^2(theta)+(2(H)/R)}^1/2+sin(theta)] Due to the earth projection

where "H" is satellite height and and R is the earth radius

My question is Can you help me to derive this equation? how they have obtained it?

Regards

**broken link removed**

 

[jeeudr]

I am not sure what theta means in your case.

For the simple situation without refraction, OAB is a right angle triangle,
\[ OA^2 = AB^2 + BO^2 \]
\[ => {(R+H)^2 = D^2 + R^2} \]
\[ => {D = \sqrt{H^2 + 2RH}} \]

 

[Rihanna]

I am not sure what theta means in your case.

For the simple situation without refraction, OAB is a right angle triangle,
\[ OA^2 = AB^2 + BO^2 \]
\[ => {(R+H)^2 = D^2 + R^2} \]
\[ => {D = \sqrt{H^2 + 2RH}} \]

Yes bro jeeudr, that is nice thought. Anyone else want to share this puzzle with me

 

[mrflibble]

Sure. If you share what theta is. Right now theta is undefined (recheck that picture), so we can conjure up any funky math we feel like.

Generally speaking (not in the mood for guessing what theta is ) you should substitute x = sin(theta), then you see it becomes a little simpler.

 

[Rihanna]

Sure. If you share what theta is. Right now theta is undefined (recheck that picture), so we can conjure up any funky math we feel like.

Generally speaking (not in the mood for guessing what theta is ) you should substitute x = sin(theta), then you see it becomes a little simpler.

Hello mrflibble and thank you for reply

This is not an accurate graph it's just arbitrary sketching to clarify my question may i check your attempt

 

[mrflibble]

My point is that your arbitrary sketch fails at clarifying the question. Theta theta what is theta there is no theta in your non-clarifying arbitrary sketch. Also, did I mention we have no clue what theta is supposed to be?

So if you can't point out what theta is supposed to be, then I suppose the generic advice to do that substitution is about it. Well, that and maybe look at what the taylor expansion of sqrt(1+x) is for x<<1, but who knows if that is relevant.

 

[jeeudr]

without knowing what theta is , we are just grasping at straws.

you can try a couple of options, and let us know how it goes (I don't feel like trying them out myself).

i. sin(theta) = D / (R + H), i.e. theta is angle AOB
ii. sin(theta) = R / (R + H), i.e. theta is angle OAB
iii. theta is angle BOC, and provides information about the refraction capability of earth. so it would be required to calculate the overall distance from A to C through B.

try to work through each, and who knows, maybe one of them may match with what you have initially provided.

 

[godfreyl]

guessing what theta is

I suspect theta is as shown in the diagram below.

Or at least Rihanna didn't dispute this when I showed it in a previous thread (this is the third thread asking exactly the same question). Then again, maybe Rihanna doesn't know what theta is either.



Disclaimer: The question relates to a flying thingie that's further away, so curvature of the earth needs to be taken into account, and the simple equation shown here can't be used.

 

[Rihanna]

OK guys i am sorry, no i've got you. I am not as you are in math here is what theta is

sin(theta)= H/ABC

As my brother godfreyl said

Really appreciate you help guys, thank you so much

 

[jeeudr]

OK guys i am sorry, no i've got you. I am not as you are in math here is what theta is

sin(theta)= H/ABC

As my brother godfreyl said

Really appreciate you help guys, thank you so much

sin(theta) should be a ratio of two sides. One from your equation is H, what is the other ? ABC ?

What does ABC mean ? isn't that what is to be calculated AC ?

 

[Rihanna]

This is what i mean

I suspect theta is as shown in the diagram below.

Or at least Rihanna didn't dispute this when I showed it in a previous thread (this is the third thread asking exactly the same question). Then again, maybe Rihanna doesn't know what theta is either.

.

 

[jeeudr]

Consider the triangle formed by satellite A, observer B and centre of earth O,

using cosine formula,

OA^2 = AB^2 + OB^2 - 2*AB*OB*cos(90+theta)

i.e. (R+H)^2 = D^2 + R^2 - 2*D*R*cos(90+theta)

I think that you should be able to solve for D from the above equation