# RF Technology and Triangles

• June 16th 2010, 10:11 AM
rpcrazy
RF Technology and Triangles
I'm pretty bad at math, I can't remember any equations, since I haven't taken it in a long time. So here's my question

You have a 90 degree triangle with C1, C2, and hypotenuse representing a house, a tower, and the ray the RF tower is shooting towards the house.

Person A Lives on the point where c1, and hypotenuse met, the tower is c2.

Measurements are 3,563 feet between the house and tower(C1), The tower is 87 feet high, and ray is 3564 long. Now I work out attenuation issues, so I need to figure out at what lenth of c1, is the height of the hypotenuse. This is to figure out if the RF ray may having hitting a 30 foot tree that's in the way. Thank for you your help.
• June 16th 2010, 10:25 AM
CaptainBlack
Quote:

Originally Posted by rpcrazy
I'm pretty bad at math, I can't remember any equations, since I haven't taken it in a long time. So here's my question

You have a 90 degree triangle with C1, C2, and hypotenuse representing a house, a tower, and the ray the RF tower is shooting towards the house.

Person A Lives on the point where c1, and hypotenuse met, the tower is c2.

Measurements are 3,563 feet between the house and tower(C1), The tower is 87 feet high, and ray is 3564 long. Now I work out attenuation issues, so I need to figure out at what lenth of c1, is the height of the hypotenuse. This is to figure out if the RF ray may having hitting a 30 foot tree that's in the way. Thank for you your help.

That is very confused, but it seems you need Pythagoras's theorem. The square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Now you just need some elementary algebra.

CB
• June 16th 2010, 11:02 AM
rpcrazy
ok so c1 is 3,563 feet long. and c2(the tower) is 87 feet, and hypotenuse is 3,564 long.

If at 3,564 feet, the length to the hypotenuse is 87 feet...what is the length of the line from the c1 to the hypotenuse at 2,500 feet?

This is essentially what i'm asking

something like 87/3,564 = ?/2,500 or something like that. I want to figure out what height the ray(or the hypotenuse) would be at a certain point in line c1. I wish i could draw pictures for you guys but I can't

http://i812.photobucket.com/albums/z...athexample.jpg

any help is appreciated
• June 16th 2010, 11:15 AM
CaptainBlack
Quote:

Originally Posted by rpcrazy
ok so c1 is 3,563 feet long. and c2(the tower) is 87 feet, and hypotenuse is 3,564 long.

If at 3,564 feet, the lenth to the hypotenuse is 87 feet...what is the length of the hypotenuse from the c1 at 2,500 feet?

This is essentially what i'm asking

something like 87/3,564 = ?/2,500 or something like that. I want to figure out what height the ray(or the hypotenuse) would be at a certain point in line c1. I wish i could draw pictures for you guys but I can't

http://i812.photobucket.com/albums/z...athexample.jpg

any help is appreciated

No, that is just as confusing.

(but at a guess you want to use similar triangles, and try labelling your diagram)

CB
• June 16th 2010, 11:24 AM
Ackbeet
I agree, CB, that the post is a bit confusing. I think the idea is something like this:

The tower is 87' tall, and the linear distance along the ground from the tower to the house is 3563'. The direct line-of-sight distance from the house to the tower is 3564'. We assume the house is 0' tall. Imagine a rod extending along the line-of-sight from the tower to the house. How far above the ground is the rod, at a point 2500' along the ground from the house toward the tower?

rpcrazy: is this a correct interpretation of your problem? If not, could you please post a word-for-word statement of the problem from a book? Thanks!
• June 16th 2010, 11:27 AM
bigwave
use the Pythagoran theorem when the info is there
still not sure if I have this understood. but assume you mean c1 to be the base the c2 the ht of the tower.

so the new hypotenuse using 2500ft as a base and the tower of 87ft

is simply

$
(2500)^2 + (87)^2 = hypotenuse^2 \Rightarrow 2501ft
$

bear in mind that this triangle is very thin, you can use ratio's to solve but much easier to use the Pythagoran theorem when the info is there
again too i may not have the correct understanding of your question
but hope this helps
• June 16th 2010, 11:29 AM
Ackbeet
I'm not sure you can use the Pythagorean theorem, because I don't think you know more than one side. You know how far from the house, but not the height (which is the target variable), or the hypotenuse.
• June 16th 2010, 11:37 AM
rpcrazy
ok i've posted a new picture

http://i812.photobucket.com/albums/z...hexample-1.jpg

I can't think of any other way how to explain it, ackbeet you've pretty much described what I want. I was wondering if there's any type of associate property I can use to figure out the height at certain points since I already know the height of the tower.
• June 16th 2010, 11:43 AM
Ackbeet
Ok, got it. You need to use similar triangles, like you were thinking in post #3. Do ratios of sides, where you know the lengths of all but the one you want. What do you get?
• June 16th 2010, 12:15 PM
rpcrazy
ok in post 3 i edited my question, i wrote down "what is the hypotenuse" but what I really want to know is what the length is, between a certain point in c1, and the hypotenuse.

This is a real life application, it's for my work. If it's not possible i'd like to know, but I think i have enough variables here. we know all the lengths of the triangle. Ackbeet describes what the situation perfectly in his first post.

Quote:

Imagine a rod extending along the line-of-sight from the tower to the house. How far above the ground is the rod, at a point 2500' along the ground from the house toward the tower?
I need to know if trees, buildings, or whatever else are potentially going to get in the way of a wave propagation.

so to reply to your question ackbeet, considering my pic link above I no longer know the hypotenuse for the "newly formed triangle".
• June 16th 2010, 12:26 PM
Ackbeet
So, what exactly is it that you know going in to the problem? I'm assuming you know the height of the tower, and you know the distance along the ground from the house to the tower, right?
• June 16th 2010, 12:26 PM
Ackbeet
Oh, and the distance at which you want to know the height of the hypotenuse as well, I'm assuming. Is this correct?
• June 16th 2010, 12:34 PM
rpcrazy
yes that's all correct. I'm able to obtain via google maps the address, and distance from the tower. from there i can obviously calculate the sort of "virtual" wave propogation in form of the hypotenuse. so basically, a, b, and c.
• June 16th 2010, 12:39 PM
Ackbeet
Yes, you can calculate the hypotenuse from the lengths of the other two sides. You can also compute your desired height using ratios of sides; you don't actually need the hypotenuse to compute the height of the line-of-sight at any particular point. So, what do you get for the height at 2500'?
• June 16th 2010, 01:33 PM
rpcrazy
63.15ft high at 2500 feet. Is that correct though?