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The picture above shows a part of a mechanical contraption exhibited in the main gallery of the MIT museum. The contraption begins with a motor (on the left) turning at 2 1 2 rpm . It is connected to a sequence of N worm-gears, which can be seen in the picture. Each worm-gear pair reduces the rotation speed by a factor of f = 5 0 . The axle of the last gear in the sequence is embedded in a concrete wall that cannot rotate at all.
Find the smallest N that will suffice to keep the exhibit running for a century.
Details
- Each wheel has approximately 50 teeth.
- The system jams when the final wheel needs to rotate by half a tooth.
The answer is 8.
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2 solutions
An awful lot of assumptions have to be made here to even make an estimate of this. Why not an answer of 7?
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Let's assume that these gears are relatively finely machined, such that the backlash at each gear is 1/100 of a revolution of the worm. Let's also assume that the motor is solar powered, and that the solar cells, motor, gears, bearings, etc will last forever. BUT, our sun will only be around for ~1,000,000,000 years or so. If the motor runs at 212 rpm for 1 109 years, it will rotate 1.1 1017 times. Getting back to the backlash, the last worm needs to rotate 1/100 revolutions to "lock-up" or break something. This means the previous worm must rotate 50/100 revolutions plus its own back-lash or 0.51 revolutions. The next worm must rotate 50 * 0.51 revolutions or 25.5 revolutions plus its own backlash or 25.51 revolutions. The backlash starts to become insignificant, and the motor revolutions required to take the slop out of the system are ~5 1016 for 12 gear sets, or ~2.5 1018 for 13 gear sets. With our solar life expectancy assumed to be 1.1*1017 revolutions of a 212 rpm motor (or 1,000,000,000 years) then we can safely say that the sun will burn out prior to the last shaft needing to rotate. do you understand now ?
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I understand the principle involved. But didn't you just now discuss the difference between "12 gear sets" and "13 gear sets", while the "correct answer" is 8?
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@Michael Mendrin – yes the "correct answer" is 8 , i discussed that bcoz i want to show the diff. created in time between even 1 gear set, that's it !
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@A Former Brilliant Member – Let me display here powers of 5 0
5
0
1
=
5
0
5
0
2
=
2
,
5
0
0
5
0
3
=
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0
0
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5
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4
=
6
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2
5
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5
=
3
1
2
,
5
0
0
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0
0
0
5
0
6
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1
5
,
6
2
5
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0
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0
0
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7
=
7
8
1
,
2
5
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0
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5
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8
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3
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9
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9
7
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6
5
6
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2
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8
8
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8
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2
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6
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Now, in one century, the motor turns about 1 4 , 8 8 5 , 6 6 0 times (including leap year days).
So, we ask ourselves, how much rotation can be tolerated in one century? With 6 gear sets, it's about 1 0 0 0 1 of a rotation. With 7 , that reduces to about \frac{1}(50,000} of a rotation. With 8 , that goes down to about \frac{1}(2,500,000} . I can tell you from my experience in the machine shop business (a long time ago), even as much as 1 0 0 0 1 "slop" in rotation is not unusual in a long chain of gears---it is very difficult to design and build an extensive chain of gears with much more precision than that.
In short, how do we decide where to draw the line?
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@Michael Mendrin – what do u intend to ask ? did u hit the like button ? ( to make this question reach to others !)
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@A Former Brilliant Member – Well of course I liked the problem, because I liked the weird chain of worm drives to nowhere. As it really is, it's actually just kind of an art sculpture, it doesn't have any practical purpose, like being part of an old-time differential analyzer machine or something. It ends with a single free block of concrete which will turn over by the end of this universe or something. It's fun to think about it.
As a matter of fact, one can end up with an extremely slow output by combining a couple of planetary drives back-to-back, if the gear ratios are chosen carefully enough. Maybe I should post something like that as a problem?
Yes, but you are still arbitrarily deciding that the "expected slop" is 1/100 of a revolution. Real gear mechanisms vary considerably in "slop", depending on a variety of things, especially including design and construction. In many cases, the engineer doesn't want the gears to be "too tight" (as for example race car transmissions), while in other cases he does (as for example astronomical telescope drives).
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@Michael Mendrin – plz take a good look at the ques sir ! it is now edited and probably explains ur question ! you are a whole lot experienced and intelligent than me ! if the ques seems good now, sorry for the inconveniences you had, any further suggestions are welcomed . thanks sir ( i did'nt see ur age uptill now)
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@A Former Brilliant Member – Yeah that's much better
Got to check this out soon!
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haven't u checked it ? i thought u were in MIT ?
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Yes. I see this museum everyday while going to my lab but never been inside. :P
Solution by Shubham explains well. For just the maths for one century:
we need to go from 212 rpm (rotations per minute) to less than 1/100 rotation per 100 years, that is to less than:
1 0 0 × 1 0 0 × 3 6 5 . 2 5 × 2 4 × 6 0 1 = 5 2 5 9 6 0 0 0 0 0 1 rpm
2 1 2 × 5 2 5 . 9 6 × 1 0 7 = 1 . 1 1 × 1 0 1 2 (approximating to 2 decimals)
with N the number of worm-gear pair needed, 5 0 N is the total reduction to compare to 1 . 1 1 × 1 0 1 2
using a calculator (did not find a better way): 5 0 7 = 7 . 8 × 1 0 1 1 a n d 5 0 8 = 3 . 9 × 1 0 1 3 (approximated to 1 decimal)
5 0 8 > 1 . 1 1 × 1 0 1 2 a n d N = 8 is minimum number of worm-gear pairs to keep exhibit running for at least (and well over) a century
The solution of the problem relies on the existence of backlash (the play between adjacent movable parts) between the gears. By assuming that the last gear pair has a backlash of, say, 1/100 of the complete rotation one can estimate how long will it take N gear pairs to create such a rotation in the before-the-last gear. By assuming that the contraption has to exist for few hundred years (or few billions of years, as assumed by Miller) one can estimate the number or gears. In the absence of backlash one would like to rely on elasticity and estimate how much torque can be applied on the last gear before it breaks. (This can be estimated from its dimensions, and the knowledge of the material from which it is made.)
Let's assume that these gears are relatively finely machined, such that the backlash at each gear is 1/100 of a revolution of the worm. Let's also assume that the motor is solar powered, and that the solar cells, motor, gears, bearings, etc will last forever. BUT, our sun will only be around for ~1,000,000,000 years or so. If the motor runs at 212 rpm for 1 × 1 0 9 years, it will rotate 1 . 1 × 1 0 1 7 times. Getting back to the backlash, the last worm needs to rotate 1/100 revolutions to "lock-up" or break something. This means the previous worm must rotate 50/100 revolutions plus its own back-lash or 0.51 revolutions. The next worm must rotate 5 0 × 0 . 5 1 revolutions or 25.5 revolutions plus its own backlash or 25.51 revolutions. The backlash starts to become insignificant, and the motor revolutions required to take the slop out of the system are approximately 5 × 1 0 1 6 for 12 gear sets, or approximately 2 . 5 × 1 0 1 8 for 13 gear sets. With our solar life expectancy assumed to be 1 . 1 × 1 0 1 7 revolutions of a 212 rpm motor (or 1,000,000,000 years) then we can safely say that the sun will burn out prior to the last shaft needing to rotate. Now you can view the entire exhibit: