<< Previous Topic | Next Topic >>Return to Index  

Trig problem

July 6 2008 at 3:30 PM
  (Login woodyj)
YFOT

Howdy everyone.  I’ve forgotten all I once knew about trigonometry long ago and would appreciate some help.

 

Situation:  There’s a commercial hot air balloon that flies low over our house most mornings and wakes us up at 6:00 – 6:30 AM.  Those things are surprisingly loud (you can hear them coming long before you can see them through the trees) and they hover directly above us for 10-15+ minutes at times.  Asides from the noise they skim along the treetops and any sudden downdraft would put them into a tree and start a huge fire.

 

Anyhow, I’m getting my ducks in a row first before I file a complaint with the FAA regional office and would appreciate your help in solving the following problem to determine the balloon’s height above ground level.

 

We can use our digital camera to take a picture through our laser range finder and document the distance from us to the balloon.  Also, my wife has a goniometer (used by physical therapists to measure people’s range of motion) and we plan to mount it on a stick and put a bubble level on the stick so we can then hold the stick level and approximate the angle to the balloon.  Thus, knowing the distance to the balloon and the angle it should be easy to use trig to solve for the height of the balloon:

  

                 C

                 /!

               /  !

             /    !

        b /      ! a

         /        !

       /          !

 A  /---------!  B

            c

 

That is, if you are standing at point (A) and the balloon is at point (C) and angle (B) is a right angle and you measure distance (b) with a laser range finder and the acute angle at point (A) with a goniometer, how to compute distance (a)?

 

Also on a related side note, I'd like to be able to compute the height I am above something.  That is, if standing at point (C) and I laser range find distance (b) and measure angle (C), how to compute distance (a)?

 

Edit:  Also, I'd like to be able to use the Windows calculator in scientific mode to do the calculation.  Therefore, if need be any answer that involves the use the cotangent, secant or cosecant (i.e., ctn, sec, or csc) needs to be transformed to use only the three functions available in the Windows calculator (i.e., sin, cos and tan). 



    
This message has been edited by woodyj on Jul 6, 2008 5:05 PM
This message has been edited by woodyj on Jul 6, 2008 4:55 PM


 
 Respond to this message   
AuthorReply
Bikerscum
(Login Bikerscum)
YC

Re: Trig problem

July 6 2008, 3:53 PM 

a^2=(b^2-c^2) You don't need to measure angle A.

 
 Respond to this message   

(Login woodyj)
YFOT

True but no cigar ...

July 6 2008, 4:38 PM 

distance (c) is an unknown!  Whereas, angle (A) is measurable.



    
This message has been edited by woodyj on Jul 6, 2008 4:40 PM


 
 Respond to this message   
Bikerscum
(Login Bikerscum)
YC

Sorry, I misunderstood.....

July 6 2008, 5:33 PM 


 
 Respond to this message   

(Login woodyj)
YFOT

Great, looks good ...

July 6 2008, 6:53 PM 

I've got to take a shower and run off to meet someone in a few minutes.  So, I haven't had a chance to do much of anything but plug in the sqrt (2) = 1.414 for side (b) and 45 deg for angle (A) and it gave an answer of 1 for side (a).  Looks like just the thing I need to solve the general problem.

Thanks much!


 
 Respond to this message   

brnchbrkr
(Login BranchBrkr)
YFOT

Re: Trig problem

July 6 2008, 7:04 PM 


 
 Respond to this message   
Stan
(Login Springer_177)
YF

Cosine and Sine functions on the calculator

July 6 2008, 10:59 PM 

Sine(<A)=a/b that yields a=b*Sine(<A)

Similarly using the Cosine function


Cosine(<C)=a/b that yields a=b*Cosine(<C)

A simple way to remember these relations is cosine = adjacent/hypotenuse
and sine = opposite/hypotenuse. With that in mind you can recognize that
side b is opposite angle A (<A) and side a is adjacent angle C (<C). The
diagram you reference is standard and helps keep definitions consistent.

So, the formulas are a=b*Sin(<A) and a=b*Cos(<C). A scientific calculator
can readily give the Sine (Sin) and Cosine (Cos) functions. Your situation sounds
really annoying and I hope the FAA may be able to help you out.

 
 Respond to this message   


(Login n6yyo1)
YF

The answers are:

July 7 2008, 11:14 AM 

Height (a) = Distance (b) x Sin A

Second question:

Height (a) = Distance b x Cos A

SOHAHTOA Sin=O/H Cos=A/H Tan=O/A I.e. multiply the hypotenuse by the Sin or Cos of the interior angle to get the Opposite or the adjacent side.

HTH

Joe Brancato <><


 
 Respond to this message   

(Login woodyj)
YFOT

Thanks for all the responses, just what I needed ...

July 8 2008, 9:35 PM 

and, yep, it's really annoying.  My wife and I are generally pretty laid back and it takes an awful lot to pi$$ off my old lady.  But, this balloon guy has managed to get her hopping mad so it's time to take care of his sorry ass.  

It's been real interesting listening in on the balloon pilot's radio transmissions: 

·       Huge ego

·      Quick temper

·      Puts on a big air of bravado about everything (I assume to put on a show for his customers)

·      Lands in tight places he shouldn't just to show off his piloting skills and tell his customers that it would be unsafe for a lesser pilot to be doing what he's doing

·      Will purposefully (during a routine non-emergency situation) land in the middle of the state highway running through town at 7-7:30 AM during the rush hour and cause traffic backups/delays

·      Very disrespectful to his chase ground crew (I would consider it hell to have to work for him)

·      Uses a GMRS/FRS radio and transmits on a channel above number 14.  Channels 1 thru 14 are FRS and don’t require a license if you transmit with ½ watt.  The channel he’s using is for GMRS only and therefore requires a license no matter what wattage you’re transmitting at.  And, I’ll lay 100 to one odds that either he or at least one of his ground crew don’t have an FCC license (because I looked it up on the FCC website and there are only a handful of people in the entire county that have GMRS licenses); and, even if he has a GMRS license, I’ll bet that at least one of his ground crew is not an immediate member of his family and therefore can’t use one his radios to transmit back to him legally on the channel he’s using).

Not the kind of guy I look forward to meeting or crossing proverbial swords with.  Sounds like he might very well be the type of guy that would think, "Screw you buddy.  So, you’re worried about fire danger from ME the great heroic pilot that I am.  If I can't land where I want (in this tiny 1/3 – ½  acre parking lot surrounded by trees and bone dry tall weeds, etc. in a park that that is marked as a special plant study zone where no plants are to be disturbed in any way) and thrill/impress my customers with my bravado in the process; and, instead have to land in some boring, safe, public land, cleared field that's several hundreds of acres in size and another ½  mi. down my flight path, I might as well take revenge and flip a lit cigarette out the window next time I drive by your place."

So, I’m hoping I can take care of the problem and remain anonymous to the guy.  But, if and when I have to approach the guy (in the most polite and respectful manner I can muster), I fully expect that he will cop an attitude with me at some point in the conversation.   And, then it will be time to drop the hammer on him:  BLM (owns the park), County Fire Prevention, County Sheriff (ideally I might start recording his radio transmissions and hope to record him when he says he going to intentionally land on the state highway in a non-emergency situation and then take a picture of him delaying traffic), FAA, FCC and BBB.  Can anyone else think of any legal dirty tricks I can also use?  Because I’d really like to nail him to the wall and I might as well leave no stone unturned while I’m at it.

Oh well.  We'll see what happens.  At least I now have an excuse to upgrade my laser range finder to one of the newer $319-$499 models with an inclinometer that gives angle as well as range to a target.  Anyone interested in my Bushnell yardage pro range finder that IIRC I paid around $230 for a couple of years ago?  It's in excellent condition.  I'm not sure where the instruction sheet is but it's not needed since it's real simple to use.



    
This message has been edited by woodyj on Jul 8, 2008 10:02 PM


 
 Respond to this message   

(Login Marmot_Militia)
YFOT

Why not just laser to the bottom of the ballon from "B"

July 9 2008, 12:08 PM 

straight up? MM

 
 Respond to this message   

(Login woodyj)
YFOT

Um-m-m, because I'm too old, out of shape, uncoordinated, slow and lazy ...

July 9 2008, 6:24 PM 

to get from point (A) to point (B) quick enough and then keep running along directly below the balloon while I try to look directly up and laser it at the same time.  I'm no longer the spring chicken that I used to be and would probably hurt myself in the process.  That sounds like a younger man's game. 

P.S.:  I had surgery a couple of weeks ago and have been in incredible pain and doped up on major painkillers since then.  Today's the first day I've felt halfway human.  Thanks again to all that read my rants and responded.


 
 Respond to this message   
Current Topic - Trig problem
  << Previous Topic | Next Topic >>Return to Index  

D I S C L A I M E R
JK's Airgun Forums, moderators, and administrators are not responsible for any problems that may occur from reading or using content posted on this forum, as they are the exclusive responsibility of two parties: the person who posted it and the person who acted on said posted information.Use of our forums from people under 18 years old, only with legal guardian(s) present.