Comments

The freewheel returns to its original position every 20 teeth, correct? So you just need a multiple of 20, which will end in a zero. There are only 2 numbers that will tive you a product that ends in zero when you multiply times 44 (a 10 and a 5) Five x 44 equals 220, so the chainwheel must turn 5 times. Maybe the constructivist would like that since I just reasoned "my own method" through this problem!!!! :)

-- CarolynMorgan - 15 Jun 2005

Oooops!! Make that "give" not "tive" !!

-- CarolynMorgan - 15 Jun 2005

Both gears are in their original positions whenever the number of teeth that have 'gone by' is a multiple of the number of teeth on the wheel. So the least number of teeth that can 'go by', leaving both gears in their original position, is their least common multiple. That number is 220. The chainring has 44 teeth, so as Carolyn says, the chainring rotates 5 times in that time.

Greetings from hot and muggy DC.

-- CarolynJohnston - 15 Jun 2005

Seems like a relatively straightforward 'least common multiple' problem. Multiply the two numbers (44 and 20) and divide by their greatest common factor (4)--a great formula for LCM, by the way, that they don't give in modern textbooks. Divide by 44 to find the number of times the chainwheel must turn to return to its original position (5). Then divide by 20 to find the number of times the flywheel must turn to return to its original position (22).

Many of these types of mechanisms are designed so that the numbers of 'teeth' in the contact wheels are relatively prime--i.e., have a GCF of 1--which reduces wear and tear.

Also a handy mathematical tool--given the right data inputs--for determining when certain planets are aligned. Something like this was probably used to determine when the best time would be to launch a rover to Mars!

-- JdFisher - 15 Jun 2005

An important question is whether the problem was posed with or without any preliminary discussion about how this type of problem can be solved. It would be nice if students could figure this out without any talk of LCM (or any other technique), as a constructivist would want, but not as the regular teaching methodology. Another question is whether the teacher reviews the proper or best methods afterwards and assigns the students (for individual work) other questions that fit this class. As I have said in the past, what is the use of constructivism if one child in the group "discovers" the solution and teaches it (perhaps poorly) to the other kids?

-- SteveH - 15 Jun 2005

I looked at this problem as a ratio. Every time the chainring turns, the freewheel turns 2.2 times. Then it is obvious that you have to multiply by 5 so that the number ends in zero. The other option is to say that every time that chainring turns, the freewheel turns 11/5 times. Multiply by 5 and you get 5 turns for the chainring and 11 turns for the freewheel.

-- AnneDwyer - 15 Jun 2005

Whoa.

OK, I'm gonna dive in here SOON.

But first, dinner.

-- CatherineJohnson - 15 Jun 2005

With respect to Steve H's comment, it would appear that a discussion of proportion and ratios would provide the proper background to solve the problem. The comments with solutions used such approach. Constructivists would look at this page and delight in how many "different ways" people solved the problem. The constructivist's conversation would end there, without pointing out that all the different ways have the same thing in common: ratio.

BG

-- BarryGarelick - 16 Jun 2005

Barry--thank you.

That's exactly right.

I still have trouble expressing this aspect of math, and it's what constructivists seem never, ever, to address.

I have in mind some future posts on this subject....

-- CatherineJohnson - 16 Jun 2005