Intersection between Involute Curve and a Circle

[Redeemed]

Homework Statement

Hello, as part of a computer science project, I have been tasked with writing an application that will generate an involute spur gear based on given criteria. To do this, I will need to be able to find the point of intersection between the involute curve (used to define the edge of each gear tooth) and a circle that can be at an arbitrary point. Below is my attempt at solving this problem. I was hoping someone might be able to give me a nudge in the correct direction, as I reach a point where I am unsure how to proceed. Any help would be appreciated, thanks in advance.

Homework Equations

Involute Curve Equation

x = A(cosT + TsinT)
y = A(sinT - TcosT)

Circle EquaTion

(x - J)^2 + (y - K)^2 = R^2

The Attempt at a Solution

Substitute involute curve equation into the circle equation and solve for T

((A(cosT + TsinT)) - J)^2 + ((A(sinT - TcosT)) - K)^2 = R^2

Expand...

(AcosT + ATsinT - J)^2 + (AsinT - ATcosT - K)^2 = R^2
(AcosT)^2 + AcosT ATsinT - J AcosT + AcosT ATsinT + (ATsinT)^2 - J ATsinT - J AcosT - J ATsinT + J^2 +
(AsinT)^2 - AsinT ATcosT - K AsinT - AsinT ATcosT + (ATcosT)^2 + K ATcosT - K AsinT + K ATcosT + K^2 = R^2

Grouping like terms...

(AcosT)^2 + 2(AcosT ATsinT) - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + (ATsinT)^2 + J^2 +
(AsinT)^2 - 2(AsinT ATcosT) - 2(K AsinT) + 2(K ATcosT) - AsinT ATcosT + (ATcosT)^2 + K^2 = R^2

(RcosT RTsinT) and (-RsinT RTcosT) cancel each other out...

(AcosT)^2 - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + (ATsinT)^2 + J^2 +
(AsinT)^2 - 2(K AsinT) - 2(K ATcosT) - AsinT ATcosT + (ATcosT)^2 + K^2 = R^2

cos^2 and sin^2 can be removed using Pythagorean identity...

cos^2T(A^2 + (AT)^2) - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + J^2 +
sin^2T(A^2 + (AT)^2) - 2(K AsinT) - 2(K ATcosT) - AsinT ATcosT + K^2 = R^2

(A^2 + (AT)^2)(cos^2T + sin^2T) - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + J^2 - 2(K AsinT) - 2(K ATcosT) - AsinT ATcosT + K^2 = R^2

(A^2 + (AT)^2)(1) - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + J^2 - 2(K AsinT) - 2(K ATcosT) - AsinT ATcosT + K^2 = R^2

At this point, I am unsure how to proceed...

(A^2 + (AT)^2) - 2(J AcosT) - 2(J ATsinT) + AcosT ATsinT + J^2 - 2(K AsinT) - 2(K ATcosT) - AsinT ATcosT + K^2 = R^2

After struggling with this for quite a while, I can't seem to find anything to do at this point that seems to be making headway toward solving the equation for T...

 

[YAV]

Hi,

Here is the solution to your problem if the involute and the intersecting circle are concentric:

Problem Statement:

Involute has base radius R, its parametric equation is: x = R*(cos(t) + t*sin(t)); y = R*(sin(t) - t*cos(t));
Circle has radius r, equation: x^2 + y^2 = r^2.

Solution:

take the square of involute equation:
x^2 = R^2 * (cos(t) + t*sin(t))^2 ; y^2 = R^2 * (sin(t) - t*cos(t))^2

add the two above equations and realize that the LHS (left hand side) is equal to r^2 (from the equation of the intersecting circle). after developping the square you get:

r^2 = R^2 * [(cos(t)^2 + t^2*sin(t)^2 + 2*t*sin(t)*cos(t)) + (sin(t)^2 + t^2*cos(t)^2 - 2*t*sin(t)*cos(t))]

then remembering that sin(t)^2 + cos(t)^2 = 1, the above equation simplifies to:

r^2 = R^2*(1+t^2)

which is easily solved for the intersecting value of the parameter t:

t = sqrt( (r/R)^2 - 1 )

Note that you can then get the coordinate of the intersection point by inserting this value of t into the parametric equation of the involute.
Also, this solution is real only if the base radius of the involute is smaller than that of the intersecting circle, which makes perfect sense.

I assume it doesn't take much from this to derive the solution when the circle and the involute are not concentric.cheers,YAV