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Wizard
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February 20th, 2019 at 9:12:38 PM permalink


The picture above was just taken at the National Museum of Mathematics in New York. As you can see, the tricycle has square wheels. I assure you the ride is smooth.

For the problem at hand, assume we have a square wheel of side length 2. It goes over a curved path such that the center of the square is always at a constant height. If the wheel makes a full revolution, how much horizontal distance will it cover?

Here are some equations you may find helpful:

cosh-1(x) = ln(x +/- (x2-1)1/2)

The derivative of cosh-1(x) = 1/(x2-1)1/2

This one is worth a beer and an appetizer. However, the rule of 24-hour old on previous winners applies.
Last edited by: Wizard on Feb 21, 2019
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
EvenBob
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February 20th, 2019 at 10:49:11 PM permalink
Quote: Wizard

The picture above was just taken at the National Museum of Mathematics



There's a museum of math? That's so
wrong on so many levels.. Is it to
mock us that failed 9th grade algebra?
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OnceDear
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AxelWolfJoemanRS
February 21st, 2019 at 3:00:44 AM permalink
Quote: Wizard



The picture above was just taken at the National Museum of Mathematics in New York. As you can see, the tricycle has square wheels. I assure you the ride is smooth.

Pah! A square wheeled unicycle would have been more impressive.
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aceofspades
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February 21st, 2019 at 6:01:47 AM permalink
It seems it would only work on that specialized track though?

Wizard
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February 21st, 2019 at 8:58:12 AM permalink
Quote: EvenBob

Is it to
mock us that failed 9th grade algebra?



Yes!
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Wizard
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February 21st, 2019 at 8:59:17 AM permalink
Quote: aceofspades

It seems it would only work on that specialized track though?



That is the point of the question -- what form would that track take?
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docegghead
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February 21st, 2019 at 2:25:40 PM permalink
I spent a bit of time trying to derive a formula for the height of the curve above the floor. The dips in the curve are at height 0 (floor height). When the square is standing on its point with the point on the floor, the center of the square is sqrt(2) above the floor. The center must always be sqrt(2) above the floor for a smooth ride.

When the square is resting at the top of the curve, the center is located 1 above the top of the curve. That means the height of the curve above the floor is sqrt(2) - 1.

I messed around with triangles of the center, to the point of contact, and to the point on the ground sqrt(2) below the center, but didn't make much progress on a generalized formula. Either way, I doubt I would have been able to get the forumula in terms of cosh, even though that was in the hint you gave.


A quick google of "square wheels math" brought up a site at Vanderbilt that got me the formula I needed.

The professor used the same dimension you did to keep the math simple. The formula for the height of the curve is:
y = sqrt(2) - cosh(x)

x = 0 is the maximum height of sqrt(2) - 1

To find the horizontal distance travelled, we need the distance to the minimum height of 0:

0 = sqrt(2) - cosh(x) =>
cosh(x) = sqrt(2) =>
x = acosh(sqrt(2)) =>
x = 0.88137


That means the center of the square travels 0.88137 to make it the 45 degrees from on point to the top of the curve. 8 of those gives a complete rotation of the square.

The horizontal distance traveled in one rotation is:
7.051
Wizard
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February 21st, 2019 at 2:41:32 PM permalink
Everything above is true, but for the beer I think you did too much cheating. I'm hoping to find a solution given only the two hints in the original post.
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Wizard
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February 22nd, 2019 at 2:37:13 AM permalink
I'm opening this up to everybody. The beer is still available for the first thorough solution.
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Wizard
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February 22nd, 2019 at 3:37:53 PM permalink
I hope you guys are busy working on it. The hanging rope problem should have been good practice.
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Wizard
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February 22nd, 2019 at 6:35:31 PM permalink
You guys are letting me down.
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RS
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February 22nd, 2019 at 9:01:36 PM permalink
Quote: Wizard

You guys are letting me down.


It’s just a tough problem. Or maybe it’s just so easy, everyone’s like, “Wow that’s too easy, I’d feel embarrassed by answering it. O.O”

Tbh I have no idea where to even start. I mean if I thought about it for like a Long time I could maybe figure out the curvature of the partial circle. Then figure out what % of a circle that is. Then figure out the “sort of diameter” thing (it’s not the real diameter because it doesn’t go through the origin, or at least not necessarily). And for some reason my phone autocorrects the word Long to have a capital L. Wth?

So that’s my detailed answer. Can someone at least confirm if the logic is sound?

Edit: Obviously you’d multiply the “sort of diameter” by 4, because that’d be one full revolution.
Wizard
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February 22nd, 2019 at 9:15:20 PM permalink
Quote: RS

Tbh I have no idea where to even start.



I think you need to start over. Here is a hint for you:

Consider the triangle with corners on the point where the square touches the path, the center of the square, and point closest to the center on the side that touches the track. Next, consider the angle at the angle of that triangle at the center of the square.

Next, consider an incremental change as the square goes down the path. Consider the angle between the point where the square touch the track, an incremental change along the x-axis and an incremental change on the y-axis. That angle will the same as the previous paragraph.

Then, equalize the secant of each angle. The solve for dy/dx. That is about as far as I'll go with this hint.
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RS
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February 22nd, 2019 at 10:43:54 PM permalink
Quote: Wizard

I think you need to start over. Here is a hint for you:

Consider the triangle with corners on the point where the square touches the path, the center of the square, and point closest to the center on the side that touches the track. Next, consider the angle at the angle of that triangle at the center of the square.

Next, consider an incremental change as the square goes down the path. Consider the angle between the point where the square touch the track, an incremental change along the x-axis and an incremental change on the y-axis. That angle will the same as the previous paragraph.

Then, equalize the secant of each angle. The solve for dy/dx. That is about as far as I'll go with this hint.


That was like the opposite of a hint. Now I’m even more confused. I haven’t done calculus for over 6-7 years now.

Once I finish this Chipotle and an episode or two of The Walking Dead, I’ll give this a shot. But I’ll do it my way, not this confusing “derivative of the secant square rooted theta axis as x approaches alpha” black magic.
Wizard
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February 23rd, 2019 at 5:23:54 AM permalink
Quote: RS

Once I finish this Chipotle and an episode or two of The Walking Dead, I’ll give this a shot. But I’ll do it my way, not this confusing “derivative of the secant square rooted theta axis as x approaches alpha” black magic.



There's more than one way to skin a cat. I can't see getting to the end without calculus though.
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unJon
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February 23rd, 2019 at 8:15:43 AM permalink
Quote: Wizard

There's more than one way to skin a cat. I can't see getting to the end without calculus though.



I was wondering if you couldn’t just use trig to solve, but am traveling and don’t have easy access to paper and pen.

Start with the square wheel laying on a side flush on flat ground with one end on the origin line. As the wheel turns on a corner, the center of the square traces a quarter circle with radius of sqrt(2) and center (0,0) from (1,1) to (-1,1). Given the formula for that circle, we now know how we need to vary the height of the ground at any point to keep the center of the square at y=1. The trick though is that as you vary the height of the ground, you no longer trace a circle with the center, but instead pinch it along the x axis because it traces that shape along the arc length of the path rather than along the x axis.
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Dalex64
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February 23rd, 2019 at 10:56:25 AM permalink
well, just kind of brainstorming here


For the square to be in contact with the circle continuously, doing a quarter rotation, the slope at each start and end of the arc has to be 45 degrees

if it is greater than 45 degrees, then the square won't fit into the valley between two arcs.

If it is less than 45 degrees, then between two arcs the square will be rotating freely between, but not along, two arcs, so I don't think there is enough information given to calculate the distance.

What distance does the bike move for each arc? That is the length of the segment along the base of the arc.

now, given an initial and final slope of 45 degrees, the pie-slice of a circle is from 45 degrees through 90 to 135 degrees. That is a quarter of a circle.

the segment length of a quarter of a circle is sqrt (r2 + r2)

ok, so what is the radius when the arc length is 2? well, that means the circumference of the circle is 8

a circle with a circumference of 8 has a radius of 1.27324

plugging that into the formula above, i get a segment length of 1.8

That is how far the bike moves as it traverses each arc/hump, or 7.2 for a full rotation of a wheel.


I've gotten these way wrong before, so no promises.
RS
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February 23rd, 2019 at 1:40:43 PM permalink

Square's sides are 2 inches each. That means the distance from one corner to the center is 1.4142 inches. The distance from the nearest point on the square to the center is 1 inch.

Therefore, the center of the wheel is 1.4142 inches above the ground, constant. And the distance from the ground to the peak of the curve is 0.4142 inches. The curve's length is 2 inches.

I looked online and found a calculator for using the arc length and chord length (2 and 0.4142 respectively), and it tells me the chord length is 1.75.

So my answer is 1.75 * 4 = 7
Wizard
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February 24th, 2019 at 3:35:39 PM permalink
Yes, the answer is somewhere between 7 and 8. Those were good estimates in the last two posts. For full credit I would like to see a full expression of the answer, without a sinh or cosh.

Does anyone need more time or should I just provide my solution?
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RS
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February 24th, 2019 at 3:57:06 PM permalink
1.7491 * 4 = 6.9964

Final answer

Not sure why the assumption should be made that the bump is 1/4 of a circle.
DRich
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February 24th, 2019 at 5:42:49 PM permalink
Wizard, did you enjoy the math museum? I enjoy basic math and statistics but clearly not at your level. That museum is on my list the next time I get to New York.
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Ace2
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February 24th, 2019 at 10:13:07 PM permalink
I didn’t know that the “track” would be a series of inverted catenaries (rope problem hint) but that seems logical. The “sag” of them will be 2^.5 - 1 = h =~.41.

This is like a catenary with poles of height h and a cable length of 2 . Cable touches ground and its “a” value is ((2/2)^2 - h^2) / 2h = 1.

The half distance between the poles is 1 * Ln (1/1 + ((1/1)^2 + 1)^.5) = Ln(1 + 2^.5) = .8814

So for each half turn on the track, the square moves horizontally by 0.8814. That times 8 half turns is a total distance of 7.051.

Just a guess.
Last edited by: Ace2 on Feb 24, 2019
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ChesterDog
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February 24th, 2019 at 11:05:42 PM permalink
Quote: Wizard

Yes, the answer is somewhere between 7 and 8. Those were good estimates in the last two posts. For full credit I would like to see a full expression of the answer, without a sinh or cosh.

Does anyone need more time or should I just provide my solution?



8 * ln( sqrt(2) + 1 ) or about 7.05099
Wizard
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February 25th, 2019 at 5:57:44 AM permalink
Congratulations, that is the right answer. You beat ChesterDog by minutes, which is interesting, considering I posted the problem four days ago. I would have preferred a more rigorous solution, but that is enough to get the beer.

Here is my solution (PDF). I welcome all comments on it.
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Wizard
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February 25th, 2019 at 6:09:19 AM permalink
Quote: DRich

Wizard, did you enjoy the math museum? I enjoy basic math and statistics but clearly not at your level. That museum is on my list the next time I get to New York.



I liked it but didn't love it. It seemed to be targeted to a very young age, like 4 to 12. There were lots of games and things to do, but I think most kids would not come away with the mathematical point behind whatever they were doing. To be honest, I didn't get the point behind a lot of the activities either. I probably would come away with a better review if I had more time, but as usual in New York, I spread myself rather thin and only had about 90 minutes there. The museum itself is rather small, I'd say about 3,000 square feet of public space. Many of the activities were difficult to understand how to control it or the object. There are lots of staff members floating about, mostly Asian males, probably college students, who were more than happy to discuss the math behind anything.

While my review is rather luke warm, I bought it up to my daughter's math teacher, who just raved about it. However, she went on a Christmas when nothing else in New York was open, so may have been grateful to do anything.
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DogHand
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February 25th, 2019 at 7:55:26 AM permalink
Quote: Wizard

Congratulations, that is the right answer. You beat ChesterDog by minutes, which is interesting, considering I posted the problem four days ago. I would have preferred a more rigorous solution, but that is enough to get the beer.

Here is my solution (PDF). I welcome all comments on it.



Wizard,

Your solution has misaligned parentheses: on the bottom of page 2 you wrote "sqrt(2 - y)", but you meant "sqrt(2) - y".

Oddly enough, a week before you posted this problem, I read about the mathematics of square-wheeled cycles in an Alex Bellos book "Alex Through the Looking-Glass".

Dog Hand

Edit: replaced ambiguous pronoun
Last edited by: DogHand on Feb 25, 2019
Wizard
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February 25th, 2019 at 9:28:42 AM permalink
Quote: DogHand

Your solution has misaligned parentheses: on the bottom of page 2 you wrote "sqrt(2 - y)", but you meant "sqrt(2) - y".



Thanks, good catch.
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Ayecarumba
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February 25th, 2019 at 4:01:10 PM permalink
Quote: Wizard

For the problem at hand, assume we have a square wheel of side length 2. It goes over a curved path such that the center of the square is always at a constant height. If the wheel makes a full revolution, how much horizontal distance will it cover?



If "it" = the total tread of the wheel, then if the square wheel never slips, won't the total horizontal distance covered by the tread = 8?

Imagine a double exposure time lapse photo of the wheel, one exposure as it hits the peak of a bump, then another as it finishes a 1/4 turn, and the wheel reaches the peak of next bump. In the photo, would the edges of the square wheel overlap, align, or have a gap?

I imagine "align", which means if I took a four exposure time lapse photo, when the wheel hit the peak of each bump, wouldn't the wheel appear as four boxes next to each other, i.e., a rectangle with sides 2 and 8?
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ThatDonGuy
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February 25th, 2019 at 4:32:47 PM permalink
I took a look at the problem, but it came down to calculating curve lengths, which is a little outside of my expertise.

I do have one question: does the problem assume that, when the square is turned so that one of the vertices extends as far down as possible, it touches the point where two of the "bumps" of the curve meet? I have a feeling the problem's solution would be different if it did not.
Wizard
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February 25th, 2019 at 5:37:34 PM permalink
Quote: ThatDonGuy

I took a look at the problem, but it came down to calculating curve lengths, which is a little outside of my expertise.



It isn't easy to stump you.

Quote:

I do have one question: does the problem assume that, when the square is turned so that one of the vertices extends as far down as possible, it touches the point where two of the "bumps" of the curve meet?



Yes.
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RS
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February 25th, 2019 at 8:40:52 PM permalink
Just so I understand — the bump isn’t a partial circle, but some elliptical shape? Is that why my answer was wrong? Like a football.
Wizard
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March 24th, 2019 at 6:21:11 PM permalink
I finally put together a video of this problem. In retrospect, it isn't one of my best, but I hope to make more videos using the white board, which was a major epic to install.



Direct: https://youtu.be/-t7xzjPz2SQ

I'm afraid to ask for comments, because I know it stinks.
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VladAlex1
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January 8th, 2023 at 4:15:31 AM permalink
you are at this clip
https://youtu.be/EbEQ86-dcMs?t=3
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VladAlex1
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January 8th, 2023 at 4:19:04 AM permalink
full episode link
https://youtu.be/EbEQ86-dcMs?t=549
I’d rather have to be a lucky player than good one.
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