Empiricle Rules

[Jackalope]

Just beating the bushes to see if anyone has equations for aspects of EM field characteristics pertaining to MD that they can share.

In theory, target signal strength decreases at the inverse sixth of the depth (round-trip). A target at 2 feet will be 1/64th as strong than at 1 ft (1/(2)^6). The inverse cube rules applies to target size, so that a target 1/2 smaller will produce a signal that is 1/8th as strong (1/(2)^3).

Together, if a target's cross-sectional area is doubled (twice as large) and has a depth that increases from 1 foot to 2 feet (100% increase or doubled), it has been stated that this would produce a signal strength that is 1/4216th as strong. However, I have not been able to determine a formula that bears out that result. It would have to account for both affects, 64x and 8x decrease in signal strength.

Any ideas?

John

 

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[EE THr]

John;

If the size of the target increased, you would use your formula to calculate an increase in signal strength, not a decrease.

The increase in depth would, as you stated, be a decrease in signal, though.

 

[Jackalope]

oops,

That should be if the target size is halved and the depth doubled, then how can the two combined effects be resolved into one formula.

The signal is not just 1/64th as strong or 1/8 as strong - it is both at once. Someone else stated somewhere that it would work out to 1/4216th as strong with both losses combined. Just don't quite see how that works. Need to define a formula that reproduces that result.

John

 

[Montauk3]

That means that it's still crackers to slip a dropsy into the rozzer mode.
If not, justify the lipstamp into 14% of the mellotones.

 

[EE THr]

John;

Using your formulas, set up a hypothetical situation as follows---

You have a coin at some amount of distance from the center of your detector coil, say 1 inch.

You have some kind of meter which measures signal strength received by your detector (the units of measure don't matter).

Your meter reads 1,000 signal strength units.

You now replace the coin with one of half the size, and read 1/8 of the original signal strength, or 125 units.

Next you move that half-sized coin to a point 2 inches away from the center of your coil, and read 1/64 of its first signal strength of 125 units, which is 1.953125 units.

To get a fractional representation of this decimal result compared to the beginning signal strength of 1000 units, divide the 1000 by 1.953125 and that equals 512 or 1/512th.

To see if this is in the right ballpark, round the 1.9 up to a 2, and note that there are 500 2s in 1,000.

 

[Jackalope]

If a signal is 1/64 as weak because it is deeper and 1/8 weaker because it is also smaller, then I'd assumed it was 1/64 x 1/8 = 1/512th weaker overall. The problem for me is that 1/512th weaker is not what another author stated, but rather 1/4216. Perhaps he's wrong but it seems probably the solution is not as simple as it appears.

John

 

[Real de Tayopa Tropical Tramp]

good morning Jackalope: The 'main' factor for field use, which no one has seemed to address, is the matrix of the soil reaction.. This is soo variable that for all practical purposes the other calculations are only good for air response, or simply said, 'crude estimations for the user'.

They are Essentially useless, except for basic air tests, not actual field conditions. Some of the other factors are the frequency, orientation, time buried allowing for an ionization of the materiel due to the earth's chemical composition, and concentration of the responsive materiel in the supposed target.

This will do for starters, don't get too wrapped up in air response, whose main use is for advertising.

Don Jose de La Mancha

 

[LX Kid]

Glad I don't have a electron spinning around loose in the conversation! May the force be with you guys!

 

[EE THr]

Jackalope;

Sometimes combinations of certain things will create a result which is very different than the sum of each of their individual results would be.

That's why I broke the problem down into two steps, so it would not be seen as an instantaneous combination, but what it actually is---two individual and separate steps. As such, if there is a different way to calculate it, I'd sure like to see it along with an explanation of why.

Also, there is what Real de Tayopa included, but that stuff would be exterior to the formula you asked about.

If you would like the opinions of people who work on and design detectors, you could ask on this forum: http://www.geotech1.com/forums/