Question about GPS Coordinates

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February 6th 2010, 05:47 PM
Sarnick

Question about GPS Coordinates

Hi all,
Just joined the site and have a question. I don't know if this is a question for the trigonometry forum but I'm not sure where to ask. I have a question about GPS coordinates (degrees, minutes and seconds). I have to find a spot and need to add 73 minutes to the minutes of the coordinates 071* 59.922 and add 10 minutes to the minutes of the coordinates 40* 58.647 to come up with the correct coordinates but I was never very good at math. I know once the minutes exceed 60 you have to add to the degrees but I'm having trouble with the conversion. Any help would be greatly appreciated.

Thanks
February 6th 2010, 06:01 PM
VonNemo19

Quote:

Originally Posted by Sarnick

Hi all,
Just joined the site and have a question. I don't know if this is a question for the trigonometry forum but I'm not sure where to ask. I have a question about GPS coordinates (degrees, minutes and seconds). I have to find a spot and need to add 73 to the minutes of the coordinates 071* 59.922 and add 10 to the minutes of the coordinates 40* 58.647 to come up with the correct coordinates but I was never very good at math. I know once the minutes exceed 60 you have to add to the degrees but I'm having trouble with the conversion. Any help would be greatly appreciated.

Thanks

What do you mean by 071* 59.922? Give it to me in word form.
February 6th 2010, 06:17 PM
Sarnick

Thanks for replying; it would be 71 degress, 59 minutes and 922 seconds West and 40 degrees, 58 minutes and 647 seconds North. The seconds don't matter for this problem.
February 6th 2010, 06:25 PM
VonNemo19

Quote:

Originally Posted by Sarnick

Thanks for replying; it would be 71 degress, 59 minutes and 922 seconds North and 40 degrees, 58 minutes and 647 seconds west. The seconds don't matter for this problem.

OK. $73'+59'=132'$ . Note that $132'\div60'=2^\circ\frac{1}{5}^\circ$

But $\frac{1}{5}^\circ=\frac{60}{5}'=12'$

Can you see where I'm going?
February 6th 2010, 06:53 PM
Sarnick

Thanks for trying to help but your confusing me more. LOL! Here's an example; If the number you had to add to the West coords was 62 then the answer would be 73 degrees, 01 minutes and 922 seconds. I was never any good at math and thanks for your time.

Does this help?
February 6th 2010, 07:06 PM
VonNemo19

Quote:

Originally Posted by Sarnick

Thanks for trying to help but your confusing me more. LOL! Here's an example; If the number you had to add to the West coords was 62 then the answer would be 73 degrees, 01 minutes and 922 seconds. I was never any good at math and thanks for your time.

Does this help?

I'll try again...(Wink)

$79^\circ{\color{red}59'}92.2''+{\color{red}73'}=79 ^\circ{\color{red}132'}92.2''$ .

Now, understand that there are $60'$ in $1^\circ$ , so we have to make the thing proper by determining how many minutes we can get out of 132 seconds.

The answer is 2 minutes with 12 seconds left over, so our final answer:

$81^\circ{\color{red}12'}92.2''+{\color{red}73'}$
February 6th 2010, 07:36 PM
Sarnick

OK so would it be 073* 12.922?

The answer is going to be 073* ?.922

81* is to high. The coords are going to be 073* something.922

How about the north coords?
February 6th 2010, 07:52 PM
VonNemo19

Quote:

Originally Posted by Sarnick

OK so would it be 073* 12.922?

The answer is going to be 073* ?.922

81* is to high. The coords are going to be 073* something.922

How about the north coords?

Oh wait, I've just read your first post again and, I'm sorry but I can see now that you meant that you wanted to add 73 minutes to 71 degrees and 59.992 minutes. This changes things.

So, we have $71^\circ59.992'+73'=71^\circ132.992'=73^\circ12.99 2'$
February 7th 2010, 04:53 AM
Sarnick

Yes I believe 073* 12.922 West and 41*.08.647 North are correct!

Anybody else??

Thank you very much VonNemo19