Motion is relative. However, "travel within the groove" might be interpreted as "relative to the groove as a fixed frame of reference", and since the groove is part of the record, that makes the record itself the frame of reference. If this were not the case, the "trick" answer to this question has been the distance the needle travels relative to the room in which the phonograph is housed, which is merely the (6 - 2 - 1) = 3-inch width (for the 6-inch radius of the record, 2-inch unusable center radius and 1-inch outer edge width) of the band that the groove makes. Strictly speaking this is not exactly true either, as the needle follows an arc of a circle centered on a pivot off to the side of the record, but for that, we'd need to know the length of the tone arm from needle to pivot. Also, it must be doubled, as both sides of a record are usually played, and strictly speaking, during the playback of the record, two grooves are followed.
So, taking the assumption that "travel within the groove" means "relative to the record as a fixed frame of reference", a close approximation can be made by assuming there are concentric circles, at 91 per usable inch of radius, rather than an archimedian spiral.
The innermost "circle" has radius 2-inches, while the outermost has radius 5 inches, so there are 3*91=273 "grooves". The average radius is 3.5 inches, for an average circumference of 7*pi inches. That gives a total length of 1911*pi inches, or about 6003.6 inches, making about 12007.2 inches for the whole record (both sides).
By the way, 1 inch is an awful wide smooth outer edge, and with only 273 "grooves" on a side, each side will last only a little over 8 minutes at 33 1/3 RPM.
Edited on March 30, 2005, 2:25 pm
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Posted by Charlie
on 2005-03-30 14:14:46 |
Non-Calculus Approximation--spoiler
