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Here’s the scrambler, a carnival ride out of my childhood:

I’m curious where my red cart will be when the ride finishes. To begin with, you might tell me a location where my red cart definitely won’t end up. Ordinarily I’d ask you what information would be helpful for you to help me answer my question, but in the interest of time, I think this may help you here:

My sense is there are a number of different ways to answer my question but I’m not sure how many. Please post your thoughts in the comments and as precise a location of the red cart as possible. I’ll update this in two days with the answer.

BTW. A follow-up question (or “sequel” in the three-act parlance): if you trace a path behind the red cart as it moves, what will the path eventually look like? (This is called a “locus” but I suppose it’s best if we postpone formal vocabulary development until our debrief.)

2013 Feb 01. Here’s the video as it runs down to the end of the ride:

And here’s a killer Desmos calculator that lets you adjust all kinds of parameters on the scrambler.

Great work in the comments. Several people analyzed the periodic nature of the scrambler. Others

21 Responses to “Let’s Do Some Math: Scrambler”

  1. on 30 Jan 2013 at 4:10 pmDaniel

    Every 3 seconds the red cart makes 1 complete rotation. In 155 second it will make 51.6 rotations. So the red cart will end up at about the 11 o’clock position still in the right quadrant where it started as the whole arm makes 31 complete rotations in that same time.

  2. on 30 Jan 2013 at 4:13 pmDavid Price

    Hi! Long time reader, first time commenter!

    So, my plan of attack would start with answering the question “Is there a time in the video at which the Scrambler is in the same overall position as it will be at the end of the ride?”

    I have a numerical answer but it’s such a good problem I don’t want to risk spoiling it for anyone if I’m correct.

    As a way of checking without spoiling, I’ll say that I’m assuming that the center of the ride is the origin and that the difference of the x-and y-coordinates of my answer is approximately 0.9019.

  3. on 30 Jan 2013 at 4:52 pmrdkpickle

    My first thought and the only one I can think is about the locus. (It’s an ellipse, right?) Blame tumblr for showing me too many cool tracing gifs like


  4. on 30 Jan 2013 at 5:08 pmRobert Kaplinsky

    I sure hope your red cart is at the bottom when it finishes. Otherwise it’s going to be a long way down.

  5. on 30 Jan 2013 at 5:32 pmPhilip McKelvey

    The red car takes 6 seconds to make a full rotation. Our ride is 2 minutes and 35 seconds… or 155 seconds. We need to find out how many full rotations the cart will make. Therefore, 155=6x [x equalling our rotations]. After dividing both sides by 6, we see that the cart will make 25.8 rotations. After the cart makes its 25 full rotations, it will only complete .8 of a rotation more. Assuming we are looking at this as a unit circle, from the 0 degree point, our cart is going to go .8 of a rotation more in the clockwise (or negative) direction. 360 degrees times -.8 is -288 degrees. Add 360 degrees to find our reference angle and we see that our cart would be at 72 degrees on our unit circle (or about midway between 12 and 1 oclock).

  6. on 30 Jan 2013 at 5:35 pmPhilip McKelvey

    the placement of the cart should be about the same as it is at 2:30:20 in the clip above

  7. on 30 Jan 2013 at 5:36 pmrdkpickle

    p.s. David – I got the same result as you… and the coordinates for location of the cart satisfy the equation I wrote for the ellipse, as well. Curious to see how you approached the problem. What a fun one!!

  8. on 31 Jan 2013 at 7:04 amTim Hartman

    Great idea for a problem. I think to hook the kids you’d need an actual video from a carnival first, right? Then you could even talk about the act of abstracting only the crucial information in moving to the next two videos.

    My guess on the elegant math way to do things would be a composition of two polar functions. Brute force looks like it could get you pretty far, though. It’s pretty easy to trace the ellipse-like shape.

    Also, what I remember being fun about the scrambler was how it felt like it sped up and slowed down while you were on it but looks like everything moving at the same speed from the outside. What an awesome calculus problem! You could graph the speed of the red car!

  9. on 31 Jan 2013 at 7:40 amDan R

    The cart will be at the same position as it was at 2:30, 2:24, 2:18, 2:12, 2:06, 2:00, 1:54, 1:48, … 0:18, 0:12, 0:06, and 0:00.

    It seems that the rate of speed the cart is traveling is not constant, but I’m going to have to do a little plotting to see what that is …

  10. on 31 Jan 2013 at 7:52 amJames McKee

    I think that perhaps your Act 2 video is off: it seems like the small cart is making a rotation every 6 seconds in Act 1, but every 3 seconds in Act 2. Can someone back me up on this one?

    Anyhow, I got the same answer as Phillip, cart makes app. 25.83 rotations in the time alloted.

    I created a simulator in Geogebra and traced the cart, and the locus is elliptical.

  11. on 31 Jan 2013 at 8:07 ama different Dave

    re: James (#10)

    In the Act 1 video, I believe the small cart is making its 3-second rotation relative to the “arm” of the larger wheel. I didn’t fully realize/appreciate this until I saw your comment, and now it’s one of the most interesting parts of the problem for me. :)

  12. on 31 Jan 2013 at 8:17 amJames McKee

    Now I see it. I can definitely see in myself the tendency to pull things out of a moving frame of reference and see them against the larger (stationary) frame of reference. Definitely and interesting aspect to the problem! When I created the Geogebra simulator, both animations were relative to the larger, stationary background, so it seemed to me that both rotations had to have the same period (6 seconds), so that’s how I viewed the problem. Thanks for the clarification!

  13. on 31 Jan 2013 at 10:02 amEileen

    Do we know how long the ride is?

  14. on 31 Jan 2013 at 12:51 pml hodge

    I really like this problem – lots of possible approaches and extensions.

    What happens if the 4 rides on the arm rotate much faster than the main body rotates? What if the main body rotates much faster than the arm rotates? The ratio of the rotation rates provided seems to be quite unique as far as the path (locus) of the red seat goes.

    If you play around with the ratio of the rates you get some very interesting paths:

    I am not 100% sure that these equations model the path of the red seat.

  15. on 01 Feb 2013 at 12:59 amDan Meyer

    Lots of interesting solutions here. I showed it to a few teachers here in Singapore and I was very surprised at how many people (like Rachel) predicted the locus was an ellipse. My intuition fails me there.

    So I added the answer video to the main page, along with a nice Desmos calculator that lets you play around with different parameters.

  16. on 01 Feb 2013 at 9:11 amDan Pearcy

    I’ve also just whipped up an applet using Geogebra.

  17. […] applet aids investigation of Dan Meyer’s Scrambler Problem. Click on the picture to go to the […]

  18. on 02 Feb 2013 at 7:56 pmMatt Vaudrey

    Has anybody mentioned the path as a polar function yet? It brings to mind the Spirographs from when we were young.

  19. on 04 Feb 2013 at 12:08 amAlec Wilson

    Similar to Dan Pearcy, I’ve made an applet in Geogebra but to show the locus of the Scrambler seat:

    Using 4 sliders and the following in the input line.

    Curve[cos(2*pi*n/V) + r*cos(2*pi*n/v),sin(2*pi*n/V)+r*sin(2*pi*t/v),n,0,t]

    Vectors & ensuing parametric equations made this quite…

  20. on 07 Feb 2013 at 4:51 amMartin Vérot

    About the problem of the overall shape obtained : This problem is quite interesting and reminds me of epicycles and the history of astronomy.

    Mathematicians used a combination of two circles with their own rotation angle to obtain various shapes.
    You can even obtain rectangles with this kind of device.

    It allowed them for example to explain the retrograde movement of Mars.

    The big idea of Kepler was to get rid of the epicycle and introduce ellipses to simplify the description of the movement with a unique and minimal description.

    To conclude : composed movements can be really counterintuitive especially with quite a lot of parameters to play with (relative angle velocity and relative ratio of the two circles).

  21. on 08 Feb 2013 at 10:34 amD MacKinnon

    Nice post and a neat topic.

    Speaking of the counterintuitive shapes that can be obtained, I was inspired to look at scrambler paths a while back:

    Sorry for the unrendered latex on those old pages – I’ll have to fix that at some point.