What's the total area I'm searching?

[+eremite]

Wandering about looking for a cache my math-challenged brain began to wonder about the total area which I need to search.

 

What I figure is that with my MeriPlats 10' radius of accuracy (best case scenario), that gives me a 314 square foot area. Right? A=pi(r2) and all that.

 

So, assuming that the placer of the cache is using a MeriPlat under a best case scenario as well, he's placing it within a 314 square foot area of where he thinks he's placing it. Right?

 

So then, what's the total area that I need to search? I mean, it's not just double (628 square feet), is it?

 

See, eremite is really mathematically challenged…

 

"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forego their use." -Galileo Galilei.

"The indoor life is the next best thing to premature burial." -Edward Abbey

[+Prime Suspect]

There's a website that shows this graphically. Maybe someone who remembers where it is could post the URL.

 

I think the search radius will be the sum of of both the hider's and seeker's error distance. So the area would be pi((r*2)^2) = 1256

 

"Don't mess with a geocacher. We know all the best places to hide a body."

[OuttaHand]

Difficult to say exactly because you have unknown variables. But here's my take on it. If the hider had the same 10' radius of accuracy that you had in that spot: If you are standing on the spot where your GPS says the coordinates are, then you could be up to 10' off in any direction. So the possible coordinates according to YOUR unit could be in a 20' diameter circle around that point. Now lets assume that the hider had the same possible 10' accuracy. He may have been standing anywhere in that 20' (diameter) circle, including anywhere along the outer edge. With his 10' of accuracy, that puts the possible location up to 10' MORE outside of the original circle. This makes your actual search area of a 20' radius from where you show the coordinates to be. Thus pi times radius squared yields approx 1257 sq.ft.

 

EDIT --- just finished typing and I see Prime Suspect must have been typing at the same time. Glad to see we both ended up with the same answer (I rounded up).

[+eremite]

Wow! Ya'll fellers is smart! Dat's wut my bruther (one o' dem enjinere, coleje edjecated types) comed up wit', too.

 

"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forego their use." -Galileo Galilei.

"The indoor life is the next best thing to premature burial." -Edward Abbey

[+Moore9KSUcats]

The above assumes that both gps receivers have the same exactly the same error in exactly the same direction. This is not a likely statistical occurance. They could just as easily had the same error in the opposite direction.

 

In engineering measurment, if the expected errors have a normal or Gaussian distribution (bell shaped curve) and are independent (two measurments where one measurment cannot effect the other) they combine in quadrature - that is the error is the square root of the sum of the squares: error(A+ = SQRT(errorA^2 + errorB^2)

 

So if your maximum error radius is 10 feet and his maximum error radius is 10 feet, the statistical mean combined error radius is SQRT (10^2 + 10^2) or 14.14 feet.

 

Now the really interesting part is if you use mean error radius as r in pi(r^2) you come up with pi(14.14 * 14.14) = 628 feet - which is the combined search area guessed at in the original post!

 

Sometimes intuition does work just as well as engineering.

[+Cachetrotters]

Thankyou, thankyou, thankyou. Stole a bunch of my thunder, but happy to see there are others here who realize the error given by the manufacturers is a statistical one, with confidence interval and level (for the others, this means error won't be any worse than stated 95% of the time, and it won't as bad as stated the majority of the time).

 

So quite right: errors combine in quadrature as long as they are at an equivalent confidence level (both 95%, both 68%, etc.).

 

don

[+eremite]

Moore (and Don)-

I'm not sure if your explanation makes me feel smarter for having guessed it correctly initially (though I was sure I was wrong); or dumber, because I have no idea what you just said!

 

But for the record, based on the fact that your formula/math resulted in the same answer as my mathematically challenged mind, I can only assume you're wrong.

 

-a confused eremite

 

"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forego their use." -Galileo Galilei.

"The indoor life is the next best thing to premature burial." -Edward Abbey

[+Cachetrotters]

You realize, of course, the penalty for exposing frauds is quite severe....

 

And a 628 sq ft area is really only a square about 25 ft on a side. Doesn't sound so big put that way.

[+Prime Suspect]

quote:

---

Originally posted by Cachetrotters:

Thankyou, thankyou, thankyou. Stole a bunch of my thunder, but happy to see there are others here who realize the error given by the manufacturers is a statistical one, with confidence interval and level (for the others, this means error won't be any worse than stated 95% of the time, and it won't as bad as stated the majority of the time).

don

---

 

From what I've read, if a Garmin unit gives an accuracy reading of, say, 15', that means there's about a 50/50 chance that the actual target will be within 15'. If you want to get up to the 95+% range, you need to double the distance given. Apparently, with Magellans, it's a little more than double.

 

"Don't mess with a geocacher. We know all the best places to hide a body."