Needle Mileage
A phonograph record has a total diameter of 12 inches. The recording itself leaves an outer margin of an inch; the diameter of the unused center of the record is 4 inches. There are an average of 90 grooves to the inch. How far the needle travel when the record is played ?
Labels: mathemagic
posted by ToM at
4:51 AM
34 Comments:
"How mar the needle travel when the record is played ?"
What is "mar"?
If this is speed, should we assume it is a 33 rpm?
"far" , my bad
9 inches out, from start to finish
i tried to work it out but got ~1*10^134 so i assume i got it worng
I dont have the time to determine the proper formula. But I see a rough summation of the circumferences of 630 concentric circles ranging from 4 in diameter to 11 in.
An approximate answer is: 11875".
I made some simplifications to get it though.
I think the exact answer might be tricky as the groove is a spiral.
wouldnt it be 3 inches?... if the diameter is 12 in. total and you subtract 4 inches for the inner circle and 2 for the outer margin, you are left with 6 total in. of actual recording. if you divide that by two because the needle only stays on one side of the record you have 3 total in. the needle will move because it only moves in one direction...towards the center of the record.
unless you mean, how long is the recorded groove on the disk?
The question is asking how long the groove is.
In my rough answer, I treated the groove as a set of circles. The difference between my answer and the true answer will be very small. My approx answer was rounded to the nearest inch.
... and don't turn the record over.
are you sure that its asking how long the groove is? technically the question just askes how far does the needle travel?... it could be a trick, especially the 90 grooves per inch is erroneous information...
roughly 5937 inches
It's not a trick question. The distance the needle travels is the same as the length of the groove (ignore lead-in/lead-out).
Also I goofed and gave twice the right length. I should have said 5937.61"
Assuming 33 rpm and 20 minutes playtime => 660 revs. As the width of the track is 3", should get 220 grooves per inch. So I guess it's a long single not an LP.
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I have now done the full spiral calculation exactly.
The answer is 5937.61" approx, which doesn't surprise me at all.
Unfortunately, the full power of calculus is required, and the resultant integral is not pretty to look at, yet alone type out here.
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I couldn't resist, the exact answer is:
(5/2)*sqrt(1+810000*pi^2) - sqrt(1+129600*pi^2)
-(arcsinh(360*pi)+arcsinh(900*pi))/(360*pi)
I hope I've typed that right. The logs I mentioned are for the indefinite integral version.
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Just to be incredibly boring, the play time of the record at 33 rpm is 8 mins 10.91 secs approx :)
Chris, the question asks how far the needle travels, correct? In order to travel, an object has to physically move. Since the needle is technically only moving sideways, toward the center of the record, I believe that it would in fact only be 3 inches. The groove that it follows is just the path of least resistance. It is basically the same as asking "you are starting at one wall. Your destination is the other wall, which is 12 ft away. If your forward stride is 40 in, and your reverse stride is 20 in, and you must take one step backward for every two steps forward, how far will you travel when you reach the other side?" The answer would be 12 ft, because the stride length and conditions of the steps is irrelevant
Last Anonymous. Your interpretation is reasonable. But if that's what the question meant, then there isn't enough information to answer it.
... you can only say the needle travelled at least 3".
Hey Chris, how did you get to your answer?? pretty crazy number.
I would start to parametrize the spiral with some sine and cosine, adapt the factors in front to get the right circumference gain per 2Pi and than integrate along the path. --> just too lazy right now
but i would be interested if you did it the same way or what was your method??
greetz
Hi t.:b:H.
I let A be the angle that the record is at, as the parameter, then let r = k*A+b, b is r at A=0 and k is chosen so that A increasing by 2*pi increases r by 1/90", and b is 2". So k = 1/(180*pi)
I did start out with A=wt etc, but realised it introduced an uneccesary variable, w.
Let s denotes distance moved, the differential distance moved:
(ds)^2 = (dr)^2 + (r*dA)^2
= (k*dA)^2+((k*A+b)*dA)^2
=> ds = Sqrt(k^2+(b+k*A)^2)*dA
Integrate with A over 0 to 540*pi, hence etc. (using 270 revs of disc).
...sadly wolframaplpha won't evaluate it.
... nor does wolframalpha
Last Anonymous. It sure would sure feel like you had walked 32'
If you you were putting an expenses claim for a business trip, using your car, from work to a place 100 miles away and back, would you say that you had travelled 0 distance and therefore wish to claim $0.00?
I should have said let s denote arc length.
Yep, now I did by myself and I get the same number, W|Alpha doesnt like that stuff very much but...
angle=t
way=w
length=L
w'=dw/dt
Parametrize the way:
w and w'
take the vector-norm of w'
||w'||
and integrate over the path:
L in our example
just wanted to show it in wolfram alpha again, to make it clear.
btw, did you solve the integral by yourself?? pretty hardcore one..
greetz
Standard integral tables on internet saved me some labour :)
Also used arcsinh[z]=ln(z+sqrt(1+z^2))
I'll probably skip it, but suspect that could do this problem with complex numbers.
Have you seen the bug on a band page, the time taken was amazing.
Pleased to see u back. Greetz
I think the trick for evaluating the integral the hard way uses a subst like A = a*sinh(y)
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I expect that I'm talking to myself, but the reason that the "trick" answer cannot be 3" is because you don't know where the tone arm is mounted. The stylus needle would have to move in a circular arc (for standard record players) - you do not know the radius or centre of the arc.
The length of the groove answers, are also incorrect unless the record is a blank disc - no allowance has been made for the side to side movement caused my any recorded content.
I'm never quite sure when people look for trick answers, if it is because they genuinely believe that answer or if it's a deliberate wind-up. Please don't stop offering them, they can be interesting. But I'd suggest the use of smilies if you want to reduce the risk of abusive responses to your suggestion(s).
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