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TRIGONOMETRY equation derivation [HELP]

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Rihanna

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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**
 

I am not sure what theta means in your case.

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

Sure. If you share what theta is. :p 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 :p) you should substitute x = sin(theta), then you see it becomes a little simpler.
 

Sure. If you share what theta is. :p 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 :p) 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 (y)
 

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.
 
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.
 
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.
 
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
 

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 ?
 

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.

.
 

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
 
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