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[DPRG] RE: more math

Subject: [DPRG] RE: more math
From: David P. Anderson dpa at io.isem.smu.edu
Date: Wed Sep 29 13:21:37 CDT 2004

>From ted at larsonland.com  Wed Sep 29 12:20:34 2004
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From: Ted Larson <ted at larsonland.com>
To: "'David P. Anderson'" <dpa at io.isem.smu.edu>,
   Ted Larson
	 <ted at larsonland.com>
Subject: RE: [DPRG] math
Date: Wed, 29 Sep 2004 10:02:31 -0700
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Whoops...typing too fast there.  You are right...360 degrees.

I didn't save my GPS coordinates...but if I go back and use the ones that
Jay Beavers posted here:
http://cylonrobot.blogspot.com/2004/09/robo-magellan-waypoints-at-seattle.ht
ml

And I pump them into my GPS equation spreadsheet I get the folllowing
result:
0.07722 kilometer = 253.3464567 feet 

My spreadsheet doesn't take into account that the radius of the earth is
different in Seattle, than it is averaged over the whole earth.  I am using
the mean earth radius of 6378.16km, so that could give me an error of up to
0.5%.

So...your result looks correct.

- Ted




-----Original Message-----
From: David P. Anderson [mailto:dpa at io.isem.smu.edu] 
Sent: Wednesday, September 29, 2004 9:17 AM
To: ted at larsonland.com
Subject: Re: [DPRG] math

Hi Ted,

What number did you get for the start to end distance?

I don't think your math is quite right below.  For an earth circumference of
40075 km, one degree is 1/360th not 1/60th. (360 degrees in a circle) and
one degree is about 111.3 km per degree, not 668 km,  That's about
365228 feet per degree, not 2,191,601.

Seismologists don't like to work in lat/lon (because the units change size
at different places on the sphere) so they use angular distance: tha angle
between the two points as projected to the center of the earth.  In angular
distance, the start and end points from last weekend's contest were
0.000699 degrees apart, hence my figure of 255 feet.  Do you have a
different number?

best
dpa


> Dave,
> 
> If you take the diameter of the earth at the equator, 12756.32 km, and 
> multiply it by PI, you get the circumference, about 40075 km.  One 
> degree is 1/60th of that, about 668 km.  One minute is 1/60th of that,
about 11 km.
> Most GPS units give their NEMA sentences in 1/10,000th increments of 
> that, which is about 1.1 meters.  I noticed with my unit, when it only 
> has a few satellites for the position fix, tends to drift about five 
> 10,000th's of a minute, which is about 4-5 meters, and you only get an 
> update about once per second.  This is why whatever solution you use 
> for navigation probably cannot use GPS alone to get to the cone, but 
> it is definitely good enough to get you within 20 feet of the cone.  
> When navigating by it alone, the drift will tend to take you on a 
> winding path too...which is interesting to watch, but less than ideal 
> for keeping on track, because then your obstacle avoidance needs to be 
> really good.  There can be lots of trees and rocks and things in a 20 foot
wide path.
> 
> So...to answer your question in feet.  One degree is 2,191,601 feet.  
> One minute is 38,089 feet.  1/10,000th of that is 3.8 feet.  Try 
> redoing your math using this, and it should work out correctly.
> 
> One of the things that bit me in the butt early on when I first 
> started doing the GPS math was that my GPS unit spews out degrees, 
> minutes only.  To convert everything to decimal degrees so everything 
> would be in the same units, I had to divide the minutes by 60, and add 
> it to the degrees.  I had originally forgotten to do this, and it really
messed up my calculations.
> It really helps to work everything out in a spreadsheet first, and 
> make sure it all works, then transfer it over to the computing 
> platform you are planning on using.  Then you can always go back to 
> your spreadsheet as a sanity check if the math isn't working out.
> 
> - Ted
>   
> 
> -----Original Message-----
> From: David P. Anderson [mailto:dpa at io.isem.smu.edu]
> Sent: Tuesday, September 28, 2004 10:21 PM
> To: dprglist at dprg.org
> Subject: [DPRG] math
> 
> Howdy
> 
> I wrote:
> 
> > That's 0.000699 angular degrees.  For earth radius of 69.172 that's
> > 69.172 * 0.000699 = .048351228 * 5280 = 255.294483849 so call it
> > 255.3 feet for last weekend's contest.  Does that sound right?
> 
> but in point of fact an earth radius of 69.172 does not sound right at
all.
> That should be,
> 
> "For an average earth radius, there are 69.172 miles per degree, so..."
> 
> Typing faster than thinking, which at this late hour is not hard.
> 
> dpa
> 
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