Ex-centrifigal Force

The short answer is; he or she would weigh less! To reach this answer, you need to take into account the fact that the earth is not a perfect sphere but an oblate spheroid, bulging slightly at the equator and being flattened at the poles. This means that someone on the equator is further from the center of the earth than at the poles and so is affected less by gravity even in the absence of rotation. I'm sure there'll be some maths to back this up.....any volunteers?

You're right about the gravitational effects. The centrifugal force on a person standing on the equator is in line with the force of gravity but opposite in direction. At the poles, the effect of centrifugal force tending to throw one off the planet is zero. So the answer depends on gravitational force - force of gravity at the equator compared to force of gravity at the poles among other things. An excellent treatise on this can be found at http://en.wikipedia.org/wiki/Gravity, by the way.

The real question is : if you took a calibrated scale to check this theory, would the temperature difference affect the calibration?

You've raised an interesting question . . . ! A scale, such as a bathroom scale, would be subjected to centrifugal force at the equator as well as the person standing on it. Would it read correctly (notwithstanding that bathroom scales have never been very acurate), given that temperature has nothing to do with the reading?

In my opinion, if the scale were a balance type (I.E. counterweights), that would be true, but a resistance type scale's measurement wouldn't be very much affected by gravity. Would it? By that I mean only the load plate would be affected and not the spring or strain guage and can be recalibrated or zeroed before applying the load. I suppose checking the calibration variation in an environment chamber prior to field measurement would be prudent. I would think that the U.S. military has already conducted this research, and it may be available through their equipment calibration studies for Arctic operations. I guess thats why we pay taxes, so we don't have to make the long trip ourselves.

A balance type scale is not affected by gravity. It is designed to measure mass. It compares the force you exert on the scale with the force exerted by the counterweight. This type of scale would read the same anywhere on earth or even the moon since the effects of gravity are exerted on you as well as the counter weights. A bathroom scale measures the force you exert on the scale by measuring the deflection your weight causes in a spring with a know force/deflection curve.

Thanks. I read at the wikipedia articles on gravity, geostationary orbits and other links. Got more confused. I was hoping there would be a simple calculation. 1/2 the diameter of the Earth is = to the moment of inertia and that has to have a fixed relationship to rotational velocity that could yield a calculated acceleration constant. And that is subtracted from the acceleration effect of gravity? I would think weighs less at the equator, but am not sure by how much.

A rough calculation based upon the rotation of the earth, the diameter, etc. shows about a 0.11 fps/sec acceleration away from the earth's core. That would work out to about 1/2 lb difference for most people. g (in fps/sec) = (v^2)/r where both velocity at the equator (7926 mi. diameter) and the radius are in feet. 0.11 fps/sec / 32 fps/sec = 0.0034 or about 0.3% affect on apparent "weight" or about 0.51 lb for a 150 lb person.

So the ratio would be about 1/300th less at the equator excluding all other effects than rotational velocity. I can't check your math, but it sounds about right. I would have thought it to be more, though.

Less! You loose a lot of weight getting to the poles. It is an arduous journey and many have perished in the attempt.

weight is a function of the distance from between two centroids of and therir respective masses. Because the earth is not spherical the distance is probably larger when you arre at the poles, hence your weight will be less.

"Because the earth is not spherical" Yes, but how much of that is due to centrifugual force at the Equator acting on the oceans? Then there's the tides. Doesn't most of the effect of gravity come from the solid iron core of the Earth?