David's Blog

Living a quiet life in Coquitlam, B.C.

Name:
Location: Coquitlam, British Columbia, Canada

Sunday, November 25, 2012

Ellipse by Cylindrical Section

Ellipses can be created in a couple ways: by passing a diagonal cutting plane through a right cylinder, or through a right cone. I plan to examine these methods in the next couple posts.

In this post, I examine the first method: creating an ellipse by taking an angular cut through a right cylinder of radius r.


When the cutting plane goes through the cylinder at an angle, the surface created takes the form of an ellipse:


To verify this statement, let's look at the cylinder from the side and affix the origin of an x-y-z coordinate system to the center line of the cylinder. Positive x is on the cutting plane, aligned with the long side of the cut. Positive y is on the cutting plane, aligned with the short side of the cut and going into the page. Positive z is perpendicular to the cutting plane.   α is the angle at which the cutting plane cut through the cylinder, with respect to the horizontal. See the following image for an illustration of these properties.


Looking at the section from above, and perpendicular to, the plane of the section, it appears as illustrated in the following image (the z-axis is coming out of the page, straight at the viewer):


(Note that the ellipse created in this way is the same as the ellipse created by projecting a circle onto a plane inclined at the same angle, α.)

Numbers will now be assigned to the section, and some numerical results calculated, to confirm that an ellipse is, indeed, described.

To allow for easy comparison with other ellipses that will be examined, the semi-major axis, a, will be assigned the value 1. The eccentricity, e, should be significant enough so that the ellipse is clearly distinguished from a circle, so it will be assigned the value 0.786151377746. These values, and other relevant quantities that result from them, follow:
a = 1
e = 0.786151377746
b = r = √(1 - e²) = 0.618033988764
α = 0.904556894284 rad

Now to compute some data points. Keep in mind that xe and ye are in the plane of the ellipse.

Start with a value of xe in the xe-ye plane.

Now to calculate the ye value at this xe point.
Projecting xe into the horizontal plane,
x = xe cos(α).
On a circle in the horizontal plane, this x value corresponds to an angle of θ = arccos(x/r).
In turn, this angle corresponds to a y-value of y = r sin(θ).
But y = ye, so the following expression for ye results:

ye = r sin[arccos((xe cos(α))/r)]

But cos(α)/r = 1/a and a = 1, so the expression reduces to

ye = r sin[arccos(xe)] = r √(1 - xe²)

But r = √(1 - e²), so the expression can be re-stated as

ye = √(1 - e²)   √(1 - xe²)

In fact, this result is the same as the standard equation of an ellipse (but rearranged):

(x² / a²) + (y² / b²) = 1

y² = b²   (1 - x²)


(Keep in mind that a = 1)
Therefore,

y = √(1 - e²)   √(1 - x²)


Apparently, a slice through a cylinder does, indeed, produce an ellipse.

Plugging some numbers into this equation,
for xe = 0.05, ye = 0.617260962835,
for xe = 0.10, ye = 0.614936054525,
and so on.

In fact, values for a quarter-section of the ellipse are tabulated in another blog post:

Blog Post: Ellipse Sample Datapoints

Datapoints computed from the expression for y above are tabulated in the column under the heading ye.
 
 

Labels: , ,

2 Comments:

Blogger Aristotelian amateur said...

David,
I'm trying to calculate how the cross section of a cylinder increases , from a circle, to an ellipse based on the angle of the cut. What's the largest area? At what angle?

3:31 PM  
Anonymous Anonymous said...

I know I am responding to a really old comment but if the cylinder is continuous in the direction of the C axis in David's diagram, the area of the cross section with approach infinity as the angle alpha approaches pi/2 radians (90 degrees). Just in case someone is curious.

6:51 PM  

Post a Comment

<< Home