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Originally posted by MrPhilNice set of ideas for sure, but he doesn't have a thermometer other than what he can touch. You are sort of on to something though. Keep thinking! If nobody gets it, I can tell the solution in one word.
My physics knowledge fails me.
I wonder if it would help to tie the calculator to the guitar string and dangle the calculator. Would spinning the calculator in circles somehow give any information to help him figure out that he is in motion?
What if he pendulums the calculator in a line with the corridor, will it achieve equal heights at both ends of the en calculate the distance.
Thats me out of ideas. I guess I'm doomed.
Phil.
Originally posted by sonhouseIs there a measurable Coriolis effect?
A guy finds himself in a room which is actually a long hallway, but he doesn't know the extent. He has with him a calculator, paper and pen, ruler( one foot and metric marks also) and a guitar string, an E string which is 0.01" in diameter (the smallest string on a guitar) and hears a voice from an unknown source saying, you are free to go if you figure out ...[text shortened]... int, he doesn't want to lose his stuff, so he has to do some thinking. What would that be?
Not sure how he will measure it though ........
Originally posted by mtthwThe coriolis effect is a 'force' even at the equator.
You're on the "equator", there won't be a Coriolis effect.
When dropping an object from a fixed point relative to he ground in a very high altitude at the direction dead centre, it will deviate from an exact vertical path according to the coriolis effect. This deviation is a function to the rotation of the system.
This is explained in the book written by Clarke in his 'Rama' books.
Originally posted by FabianFnasFair enough - that's not really what is usually considered the Coriolis effect, but as a generalisation it's the same thing.
The coriolis effect is a 'force' even at the equator.
When dropping an object from a fixed point relative to he ground in a very high altitude at the direction dead centre, it will deviate from an exact vertical path according to the coriolis effect. This deviation is a function to the rotation of the system.
This is explained in the book written by Clarke in his 'Rama' books.
Originally posted by mtthwSo by measuring the coriolis effect then you'll know the distance from the center.
Fair enough - that's not really what is usually considered the Coriolis effect, but as a generalisation it's the same thing.
Do like this: Hold a marble two meters up from the floor. Let it go. Measure the distance from the position where it should hit the floor without the coriolis effect, and the position where it actually hit the flor. This distance, in some sense, is a metric of the coriolis effect in this rotating system.
But by doing this by a shaky hand it's to crude to get any information out of it...