I'm tryin' ta get a handle on this visualisation problem..?

Question

You have a square cam, and an accentric,circular cam - bulging from the shaft. If you traced the up and down (SHM?) movement of the follower against a moving strip of paper, in what respects would the traces differ from each other, seeing that one cam is sharp and pointy and the other is smooth and circular. In both cases there is a regular changing of a ('radius' ?), but how would you describe this mathematically..?

Answer 1

the square cam would produce more of a triangle wave form where the eccentric cam would produce more of a sine wave form

Answer 2

Mark S is correct: The follower on the square cam will trace a series of slightly distorted triangles and the follower on the circular cam will trace a distorted sinusoid function.

The way to solve this mathematically is to take the camshaft axis as the origin of an x-y graph and write the functions for the edges of the cam shapes as functions of x and y. Let E be the length of the edge of the square cam. Let R be the radius of the circular cam and let F be the offset of the circular cam. (The center of the circlular cam is distance F away from the camshaft axis.) Then the equations for the shapes are converted from x-y rectilinear coordinates to r-θ polar coordinates. (Little) r is the variable radius, the distance from the camshaft axis to the point where the follower touches the cam when the cam has turned through angle θ (theta). The tracing will then be r(varies as camshaft turns) + length of follower (a constant).

Square Cam

From here I first solve for the square cam tracing.

The f(x,y) formula for the square cam has to be expressed as a piecewise function, over the 4 quarters of a cycle. Over 1st quarter-cycle f(x,y)={x=E/2, y = (E/2) tan(θ)}. Over 2nd quarter-cycle f(x,y)={x = (E/2)/tan(θ), y=E/2, }. Over 3rd quarter-cycle f(x,y)={x = -E/2, y = (E/2) tan(θ)}. Over the 4th quarter-cycle f(x,y)={x = (-E/2)/tan(θ), y=-E/2}

Next convert to r-θ system: The polar conversion formulas for converting from x-y coordinates to r-θ coordinates are x = r cos(θ) and y = r sin(θ). After a bit of algebra the result is as follows:

r = (E/2)/cos(θ) over cycle quarters 1 and 3,
and r = (E/2)/sin(θ) over cycle quarters 2 and 4.
(One can also use functions r = (E/2) sec(θ) and r = (E/2) csc(θ).)

You can then get a visualization by looking up the trig functions above, or better, by calculating them in a spreadsheet program and having the program plot them, over the 4 quarters of a complete cycle. The result will be a series of 4 triangles, side by side, each triangle with a base of 1/4 cycle and a height of (E/2) (1-1/√2), the difference between a line from axis to flat and to corner of square. The sides of the triangle will be slightly concave inward, due to the inverse sine and cosine function.

Another way to visualize the square cam tracing is to line up 4 wheels. Set the wheels side-by-side to touch each other, all in a row. Hold a large piece of cardboard over the tops of the wheels, so that only the bottom halves of the wheels are visible. The rims (outer edges) of the wheels then form 3 peaks plus 2 half-peaks (at the ends). The curve shown by these rims is very close to the tracing the follower would make as the square cam rotates one cycle. (To put this at the same scale as the tracing, the wheel diameters should be E (1-1/√2), where E is cam edge.)

Offset Circular Cam

The follower tracing for the offset circular cam can be solved in a similar way, and this time the clumsiness a of piece-wise function is avoided. Let R be the radius of the circular cam and let F be the offset of the circular cam. The x-y function for the shape is (x-F)^2 + y^2 = R^2 (the formula for a circle with center at point F on the x-axis). Again using the polar conversion formulas, the equation for r as a function of θ is obtained:

r = -Fcos(θ) + Sqrt [ F^2 cos^2(θ) + (R^2 - F^2) ].

Reminder: r = distance from camshaft axis to where follower touches cam, r depends on F, and r generally varies as the cam rotates. R is a constant, the radius of the circular cam itself, i.e. half the diameter of the circular cam shape.

If F=0 (meaning the circular cam is centered on the shaft), then obviously r = R, and indeed r in the function above reduces to R, so r is a contstant, and the follower stays at same level, not going up and down. This of course is not really a cam. As F increases slightly from 0 the follower starts to go up and down, making nearly a perfect sinusoid, but with small amplitude. As F increases more, the amplitude of the sinusoid tracing increases, but so does the amount of distortion, and the more offset, the more distortion. But with practical cam offsets, the tracing would look essentially like a sinusoid to the unaided eye.

It would take too much room to show the full derivation of the formulas here. Too bad I cannot show the graphs, either.