Mechanical Reasoning 101

[Jonathan]

 

The pins are rigidly affixed to the rod. They cannot slide along the rod, nor can it slide through them. The crayon is held by a cylinder welded to the end of the rod. One pin is closer to the crayon than the other.

The pins are mounted to the sleds via the sleds’ collars. The pins are limited to rotating in the collars on their vertical axes.

Grooves X and Y are perpendicular to each other. Sled 1 is limited to sliding in groove X, and sled 2 is limited to sliding in groove Y.

The crayon is in constant contact with the surface beneath its tip.

When horizontal force is applied to the crayon’s holding cylinder, perpendicular to the rod, what shape will be created by the crayon’s path once it has completed a full cycle?

[Jon Letendre]

32 minutes ago, Jonathan said:

 

The pins are rigidly affixed to the rod. They cannot slide along the rod, nor can it slide through them. The crayon is held by a cylinder welded to the end of the rod. One pin is closer to the crayon than the other.

The pins are mounted to the sleds via the sleds’ collars. The pins are limited to rotating in the collars on their vertical axes.

Grooves X and Y are perpendicular to each other. Sled 1 is limited to sliding in groove X, and sled 2 is limited to sliding in groove Y.

The crayon is in constant contact with the surface beneath its tip.

When horizontal force is applied to the crayon’s holding cylinder, perpendicular to the rod, what shape will be created by the crayon’s path once it has completed a full cycle?

The shape of an eye, like in the CBS logo, just the outline, of the eye. But with roundy edges on left and right, instead of pointy.

[Jon Letendre]

I rounded off the points on the left and right of the CBS logo to show what shape I mean.

Oriented so the Y groove runs left to right through the middle of the eye, the crayon currently drawing the bottom right of the eye. ... 

[Mark]

The mechanism is deceptively symmetrical.  At first glance it looks the same left and right (minus X and X) as top and bottom (Y and minus Y), but on closer examination these two aspects are different.

Let L = length of the full rod, from pin 1 to crayon.
Let M = length of the part of the rod between pin 1 and pin 2.
Let x, y = the coordinates of the crayon using the grooves as axes as in the picture.
Let x0 = the length of the projection of the “M” part of the rod onto the X axis.

Two relations of these quantities can be seen by considering certain right triangles.  The first follows from the Pythagorean theorem with rod as hypotenuse, the second from similar triangles:

(1) The square of x + x0 plus the square of y equals the square of L.
(2) x + x0  equals  x0 times L / M.

First solve (2) for x0 in terms of x then substitute it into (1) to get an equation involving only x, y and the constants L and M. Doing this and simplifying, you get an equation for an ellipse:

A squared times x squared + B squared times y squared = C squared

where

A = L
B = L – M
C = the product of A and B

Since A > B the ellipse is stretching along the Y axis.

 

[Max]

y₁² + x₂² = M²

x₃/x₂ = (M + L)/M    x₃ = ((M + L)/M) x₂

(y₃ - y₁)/y₃ = (M + L)/L     y₃ = (-L/M) * y₁

x₃²/((M+L)²/M²) + y₃²/(L²/M²) = y₁²  + x₂² = M²

which is indeed the equation of an ellipse. As the denominator in the x-term is larger than the denominator in the y-term, the long axis of the ellipse is in the x-direction.

Edited January 4, 2019 by Max
corrected typo

[BaalChatzaf]

17 hours ago, Jon Letendre said:

The shape of an eye, like in the CBS logo, just the outline, of the eye. But with roundy edges on left and right, instead of pointy.

It is called an ellipse. 

[Jonathan]

Correctamundo, ya'all. The shape is an ellipse.

J

[Jon Letendre]

10 hours ago, Mark said:

The mechanism is deceptively symmetrical.  At first glance it looks the same left and right (minus X and X) as top and bottom (Y and minus Y), but on closer examination these two aspects are different.

Let L = length of the full rod, from pin 1 to crayon.
Let M = length of the part of the rod between pin 1 and pin 2.
Let x, y = the coordinates of the crayon using the grooves as axes as in the picture.
Let x0 = the length of the projection of the “M” part of the rod onto the X axis.

Two relations of these quantities can be seen by considering certain right triangles.  The first follows from the Pythagorean theorem with rod as hypotenuse, the second from similar triangles:

(1) The square of x + x0 plus the square of y equals the square of L.
(2) x + x0  equals  x0 times L / M.

First solve (2) for x0 in terms of x then substitute it into (2) to get an equation involving only x, y and the constants L and M. Doing this and simplifying, you get an equation for an ellipse:

A squared times x squared + B squared times y squared = C squared

where

A = L
B = L – M
C = the product of A and B

Since A > B the ellipse will be stretched along the Y axis.

 

An ellipse drawing device. Nice catch, Mark.

[Darrell Hougen]

Since I came late to the party, I decided to try to solve the problem without looking at other solutions and came up with a slightly different approach.

Let r be the distance between the two pins and let the origin of the coordinate system be in the middle of the figure. Then, by the Pythagorean theorem,

x² + y² = r²

That is also the equation of a circle. We can also give the equations of a circle in parametric form:

x = r*cos(t)

y = r*sin(t)

Given that, what would the equations of the crayon be? 

Let R be the distance from the y-pin to the crayon. Then, if X and Y are the coordinates of the crayon:

X = -R*cos(t)

Y = (R + r)*sin(t)

But, those are just the parametric equations of an ellipse.

The equation of the ellipse can be written in standard form by combining the two equations:

X²/R² + Y²/(R+r)² = 1.

So, the figure is an ellipse with y as the long axis.

 

Darrell

[Jonathan]

Thanks, Darrell, and thanks, Jon, Mark and Max, for sharing how your mathy minds work.

J

[Jon Letendre]

2 hours ago, Jonathan said:

Thanks, Darrell, and thanks, Jon, Mark and Max, for sharing how your mathy minds work.

J

Thanks but I used only my mechy mind on this one.

I am shocked at how much basic algebra and trig I have forgotten. It is pissing me off, so I bought a few textbooks in the last few weeks!

[Darrell Hougen]

On 1/19/2019 at 7:27 AM, Jonathan said:

Thanks, Darrell, and thanks, Jon, Mark and Max, for sharing how your mathy minds work.

J

Just as a point of interest, someone in my family used to have a contraption like the one depicted above, except that instead of having a crayon, the end of the rod had a handle. I think it was called a "Do nothing." You could turn the crank for hours and by doing so you would accomplish ... nothing.

Darrell

[Darrell Hougen]

If you're bored and have extra time, here is a video explaining how to build one.

Darrell

 

[Jonathan]

The slightly more complex, adjustable variation isn't a "do-nothing," but a nifty router jig:

J

[BaalChatzaf]

On 1/3/2019 at 5:47 PM, Jonathan said:

 

The pins are rigidly affixed to the rod. They cannot slide along the rod, nor can it slide through them. The crayon is held by a cylinder welded to the end of the rod. One pin is closer to the crayon than the other.

The pins are mounted to the sleds via the sleds’ collars. The pins are limited to rotating in the collars on their vertical axes.

Grooves X and Y are perpendicular to each other. Sled 1 is limited to sliding in groove X, and sled 2 is limited to sliding in groove Y.

The crayon is in constant contact with the surface beneath its tip.

When horizontal force is applied to the crayon’s holding cylinder, perpendicular to the rod, what shape will be created by the crayon’s path once it has completed a full cycle?

The same thing can be done with a string and two thumbtacks.   Please see 

 

[Jonathan]

49 minutes ago, BaalChatzaf said:

The same thing can be done with a string and two thumbtacks.   Please see 

 

Yes, and there's also a method using just a strip of paper, but both the string and the strip methods are harder to control if you really need accuracy (notice the jaggy mess-ups in the video above). Also, the adjustable track jig method, like the example in the Rockler video that I posted, is SOOOO much easier for creating the precise size of ellipse that you want -- if you want, say, an 18 x 24 ellipse, you just set the pins at 18 and 24, where figuring out the same with the focal points and string method is more complex. Notice that the video that you posted doesn't address how to create an ellipse of a predetermined height and width.

J