David's Blog

Living a quiet life in Coquitlam, B.C.

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Sunday, November 25, 2012

Ellipse by Conical Section

This post examines elliptical sections which are created by taking a diagonal cutting plane through a right cone.

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

To verify this statement, let's look at the cone from the side and affix the origin of an x-y-z coordinate system to the center line of the cone. 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 cone, 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):

One feature of this section should be noted: the center of the ellipse does not coincide with the center line of the cone that produced it. The center line of the cone intersects the elliptical plane at the origin of the X-Y-Z coordinate system. HOWEVER, the center of the ellipse is at the origin of the Xe-Ye-Ze coordinate system, which is offset by an amount ro from the origin of the X-Y-Z system in the positive X direction (see the image above).

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

Values for a and e will be the same as were used in the previous post (for a section created through a cylinder):

a = 1
e = 0.786151377746

An infinite number of cone and cutting plane combinations could produce an ellipse with this eccentricity, so one more constraining parameter, gamma, will be defined:

γ = h/r

where h is the height of the cone, and
r is the radius of the bottom of the cone.

γ now serves as a parameter to define a cone uniquely by its height/radius ratio. Then

α =   arcsin[(γ e)/√(γ² + 1)]   and

ro =   a   (tan(α)/γ)

For the purpose of the following numerical example, γ = 1.27201964953.
The necessary cutting plane is then α = 0.666239432494 radians, and
ro = 0.618033988735.

To summarize:

γ = 1.27201964953
e = 0.786151377746
a = 1
ro = 0.618033988735
b = √(1 - e²) = 0.618033988764

(Note that the angle of the cutting plane, α, is much smaller in this case than it was when a section was taken through a cylinder--to produce an ellipse of the same eccentricity.)

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

Calculate the ye value at this xe point.

Projecting xe into the horizontal plane and measuring with respect to the cone axis,

x = (ro + xe) cos(α)

The radius of the cone at this level is

r = (γ cos(α) + sin(α) xe)/γ

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 ye value:

ye = r sin(θ)

As a quick check, at xe = 0,

r = cos(α) = 0.786151377757
θ = arccos(tan(α)/γ) = arccos(0.618033988754) = 0.904556894297

Then ye = r sin(θ) = 0.786151377757*sin(0.904556894297) = 0.618033988738

Remember that this value of ye corresponds to b, the mid-point of the ellipse. And because this value is the same as the one calculated from the standard equation for an ellipse, b = √(1 - e²), this is a good indication that the section is an ellipse.

An alternative to this rather roundabout way is to use the following equation for ye:

Whichever technique is used, computing some values for ye,
for xe = 0.05, ye = 0.617260962835
for xe = 0.10, ye = 0.614936054525
and so on.

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 ye above are tabulated in the column under the heading ye. Note that these values are the same as the values computed from the standard equation of an ellipse.

Apparently, taking a diagonal cutting plane through a right cone does, indeed, produce an ellipse.

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

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

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.

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Sunday, November 04, 2012

Keplers Equation: Eccentric Anomaly Values when sin(E) = sin(M)

In this post I present some information about the solution to Kepler’s Equation of Elliptical Motion (KE) at the point at which sin(E) = sin(M). This point is significant because it is the only point at which the value of sin(E) might be found which is a linear expression in time. Plots of E and M intersect only at t = 0, π, and 2π. However, plots of sin(E) and sin(M) intersect at one more point, and because sin(M) operates on an operand that is linear in time, the value for sin(E) at this time would also be known. I will refer to M and E at this point as MsE=sM and EsE=sM, respectively.

According to KE, E = M + e sin(E)
Since sin(EsE=sM) = sin(MsE=sM), this equation can be re-written as follows:

EsE=sM = MsE=sM + e sin(MsE=sM)

The only time at which this condition holds is when E and M are symmetric around π/2. Also at this time, M + E = π.

STATEMENT 1

EsE=sM - π/2 = π/2 – MsE=sM
= (e/2) sin(MsE=sM) = (e/2) sin(EsE=sM)

From STATEMENT 1, expressions can be made in terms of either EsE=sM or MsE=sM:

EsE=sM - (e/2) sin(EsE=sM) = π/2         EQUATION 1
MsE=sM + (e/2) sin(MsE=sM) = π/2       EQUATION 2

EQUATION 1 is very similar to KE itself, and can be solved numerically. A plot of sin(EsE=sM) is shown in the graph below.

From a diagram of this geometry, note that

cos((e/2) sin(MsE=sM)) = sin(MsE=sM)
sin((e/2) sin(MsE=sM)) = cos(MsE=sM)

The half-angle identities could now be used:

cos(B/2) = √((1 + cos(B))/2)
sin(B/2) = √((1 - cos(B))/2)

Therefore,

cos((e/2) sin(MsE=sM)) = √((1 + cos(e sin(MsE=sM)))/2)
sin((e/2) sin(MsE=sM)) = √((1 - cos(e sin(MsE=sM)))/2)
respectively.

These forms don’t really help, because they are even more complicated equations to solve.
At this point, the most concise expression for sin(EsE=sM) is as the solution to the equation

cos[(e/2) θ] = θ, where θ = sin(EsE=sM).

This equation is analogous to the expression for sin(ET/4), which is as the solution to the equation

cos[e θ] = θ, where θ = sin(ET/4)

Some numerical data for e = 0.786151377748:
rη = 0.792451718158,
sin(EsE=sM) = 0.933439441046,
cos(MsE=sM) = 0.358735013494,
√(1 - e²) sin(EsE=sM) = 0.576897301019

Also, on the graph below, note that a plot of cos(0.45e) looks very close to a plot of sin(EsE=sM).

Some notes for the following image:
i) sin(ET/4) is labeled sET4;
ii) sin(EsE=sM) is labeled sqEM;
iii) c1e = cos(0.45e);
iv) as e goes to 1, the equation goes to cos(θ/2) = θ,
where θ = sin(MsE=sM). This equation is solved numerically:
sin(MsE=sM) = 0.900367222589,
cos(MsE=sM) = 0.435130859037,