A perfect circle is supported by four legs, 50m long. The circle is flat when all four legs are 50m long, and tilted when the legs are different in length. The four legs are spread evenly 90 degrees apart at 0, 90, 180, and 270. Find the high and low points in degree and height when the legs are 55m at 0 degree, 70m at 90 degree, 45m at 180 degree, and 30m at 270 degree.
The radius of the circle is 100m.
Note: Since no one else responded, I'll make an attempt, though I haven’t done problems like this in quite awhile. Hopefully, the “brighter minds” on the forum can check/validate my assumptions and work. :?
Vector math can be used to solve this problem, but since this is posted in Geom and Trig, I’ll use as little as possible, not knowing what methods you have available.
One approach is to find the equation of the circle in terms of x, y, and z, and find the maximum and minumum z values. Another involves finding the angle between two planes.
If our circle were lying in the xy-plane, the equation would be x^2 + y^2 = r^2, where r is the radius.
The equation of a circle in 3-D is the combination of two equations, such as
(x-h)^2 + (y-k)^2 + (z-j)^2 = r^2
ax + by + cz = d
The first equation defines a sphere, and the second equation defines a plane intersecting that sphere. [Note: a, b, and c are actually vector directions. An alternative method is to use a system of parametric equations.] The points of intersection form a circle.
Let our circle be centered on the z-axis and raised above the xy-plane (by the amounts shown as legs) in the positive z direction. Let 0 degrees be the positive x direction and 90 degrees be the positive y direction.
The center of the circle can be found by averaging either the 90 and 270 legs ((70+30)/2) or the 0 and 180 legs ((55+45)/2) to be 50. Therefore, our first equation takes the form
x^2 + y^2 + (z-50)^2 = 100^2
To find the second equation, we’ll need coordinates of points on the circle, which we can find using the distance formula/Pythagorean theorem.
0 degree point, leg 55:
100 = [(x-0)^2 + (55-50)^2]^(1/2)
x^2 = 100^2 – 5^2
x = (9975)^.5
So, 0 degree point is ((9975)^.5, 0, 55).
By similar approach for the 90 degree point, leg 70:
100 = [(y-0)^2 + (70-50)^2]^(1/2)
y^2 = 100^2 – 20^2
y = (9600)^.5
90 degree point is (0, (9600)^.5, 70).
At 180 degrees, leg 45:
100 = [(x-0)^2 + (50-45)^2]^(1/2)
x^2 = 100^2 – 5^2
x = (9975)^.5
180 degree point is ((9975)^.5, 0, 45).
270 degree point, leg 30:
100 = [(y-0)^2 + (50-30)^2]^(1/2)
y^2 = 100^2 – 20^2
y = (9600)^.5
90 degree point is (0, (9600)^.5, 30).
Our four points are
((9975)^.5, 0, 55)
(0, (9600)^.5, 70)
((9975)^.5, 0, 45)
(0, (9600)^.5, 30)
We also know a fifth point in the plane of the circle (but not part of the circle), the center, (0,0,50). Any three known points can be used to define a plane. Consider the general equation ax + by + cz = 1 ( i.e., d = 1).
Plugging in (0,0,50) we get
a(0) + b(0) + c(50) = 1
Similarly,
a((9975)^.5) + b(0) + c(55) = 1
a((9975)^.5) + b(0) + c(45) = 1
a(0) + b(9600)^.5) + c(70) = 1
a(0) + b((9600)^.5) + c(30) = 1
Choose three (non-colinear) points/equations and solve (using matrices, for example). I got (approximately)
-.00100125x - .00408248y + (.02)z = 1
Our circle is defined by the following system of equations:
x^2 + y^2 + (z-50)^2 = 100^2
-.00100125x - .00408248y + (.02)z = 1
Next find the angle between the plane of the circle and the xy-plane:
http://mathforum.org/dr.math/faq/formul ... hreeplanes
The dihedral angle between two planes is equal to the angle between their normal directions. If the planes are given by
A1x + B1y + C1z + D1 = 0,
A2x + B2y + C2z + D2 = 0,
then the dihedral angle between them is
arccos([A1A2 + B1B2 + C1C2]/[sqrt(A12 + B12 + C12)sqrt(A22 + B22+C22)]).
Solving (using a=b=0 and c = 1 for the xy-plane and the previously calculated values for the circle plane), I get angle = .207158
Using simple trig to solve for opposite side of a triangle with hypotenuse equal to 100:
h = 100sin(.207158) = 20.568
Add this number to 50 to get the high point of the circle, and subtract it from 50 to get the low point.
Hope that helps.
