• Quick note - the problem with Youtube videos not embedding on the forum appears to have been fixed, thanks to ZiprHead. If you do still see problems let me know.

Model for Predicting Where Earrings Fall?

T'ai Chi

Penultimate Amazing
T
Joined
May 20, 2003
Messages
11,219
Some of my female relatives and friends are always dropping their earring backs and asking me to find them! I have a pretty good track record of finding them so far, but thought it would be easier if I had some mathematical model to estimate a 'search radius'.

This radius is obviously determined by the friction of the floor (ie. a deep rug vs. concrete) and the drop angle (something dropped at 45degrees will go further than something dropped for directly above).

Two questions:

1. can anyone think of anything else besides the surface friction, f, and the drop angle, a?

2. does anyone have any proposals for rudimentary models? I'm working on one from more of a statistical standpoint, but am wondering if there are good mathematical ones out there already.
 
Posibley mass of earing and hight of drop should be includid in the formular?

(search area)K= [{(90-a)+c}*(1/f)*height*mass]+c
 
How about earring friction? And earring elasticity (imagine a rubberball earring)? Earth gravity (surely you want your model to work on the moon and in outer space)? Air pressure?

Some other things you should take into account: wind, magnetic fields, animals or little children could transport the earring farther away.

Also you should take into account how tilted the floor is.
 
Don't forget friction and elasticity upon contact with the ground. An earring dropped in a shag carpet will stay right there, but if dropped on linoleum it may bounce and skitter away, even all the way across a room.
 
great so now I getting a formular along the lines of:

(search area)K= [{(90-a)+c}*{(1/f)(height*mass*g)}*{(height*mass*g)*earing elastitly}*{(tilt of floor)*(1/f)}*(air pessuer*glide ratio)]+c

And that is before we include wind, small children and magnetic fields
 
Perhaps with some efforts we could make this the long searched for Theory Of Everything.
 
Because they almost always drop just after taking them off, or when putting them in, I was thinking the height to be constant, and to be about the average female ear height. Ultimately though, it would be nice to have the height be a variable.
 
I don't remember ever having to look for my wife's earrings, and she loses everything imaginable. Much more useful for both of us would be a model for finding dropped contact lens. Sometimes they bounce, sometimes they stick, no sound to clue you in, almost invisible... many problems.
 
my mothers earings were ripped off her head while riding Disney World's Space Mountain. they were found embedded in the seat cushion after the ride.
 
geni said:
great so now I getting a formular along the lines of:

(search area)K= [{(90-a)+c}*{(1/f)(height*mass*g)}*{(height*mass*g)*earing elastitly}*{(tilt of floor)*(1/f)}*(air pessuer*glide ratio)]+c

And that is before we include wind, small children and magnetic fields
Click to expand...


Just get two long rods....
 
Assuming many earings are dropped while the wearer is engaged in some sort of physical movement, I think you need to add in velocity as a variable.
 
Ursa Major said:
Assuming many earings are dropped while the wearer is engaged in some sort of physical movement, I think you need to add in velocity as a variable.
Click to expand...

so now we have

(search area)K= [{(90-a)+c}*{(1/f)(height*mass*g)}*{(height*mass*g)*earing elastitly}{initial V*(height*mass*g)}*{(tilt of floor)*(1/f)}*(air pessuer*glide ratio)]+c

Me thinks it would be easyer to look for the earing
 
3D Vector Calculus can solve almost anything!

(Although one might reason that you are simply "thinking too much about 'stuff'"...)
 
You must log in or register to reply here.

Back
Top Bottom