Alright, so with the corrected groove length this should work out. I found this article on how to calculate a spiral length that says that this is the formula
L = Pi * R * (D+d) / 2
where L is the total length, R is the number of revolutions in the spiral, D is the outer diameter & d is the inner diameter.
So first we'd need to find how many rotations the record has. So according to pg1067's article, the outermost recording groove is placed at an 11.5" diameter for 12" records, while I'm assuming the innermost groove depends on the recording length, so let's just put it at 3.875".
So that'd mean that the total diameter would be 7.625", which is a radius of 3.8125". If we assume the lowest possible groove length as 0.04mm, we'd still have to divide by 0.08mm to get the spaces in between them. So that's 3.8125 / 0.00314961 = 1210 maximum rotations (rounding down).
So then the total length is L = Pi * 1210 * (11.5 + 3.875) / 2
L = Pi * 1210 * 15.375 / 2
L = 29222.7 inches (1dp)
So now we have to find out how fast the needle is moving along the groove.
If a record is 33 1/3 RPM, that means that the whole record is rotating at 33 1/3 RPM, meaning that it's the total circumference of the record that that speed relates to (though I might be wrong on this).
So the length of the circumference = Pi * d = Pi * 12 = 37.7 inches (1dp)
So the needle speed is 33 1/3 * 37.7 inches per minute
That's 1256 2/3 inches per minute
which is 20.94 inches per second (2dp)
29222.7 inches / 20.94 inches per second
= 1395.5 seconds
= 23 minutes 15.5 seconds
That makes more sense