- #1
cpman
- 20
- 2
Hello.
So, I'm designing an equatorial platform mount for my telescope at the moment. I'm also going to use it for another telescope that I'm in the process of building.
I know that for both of the bearings, I can use small sections of two circles cut from a cone with an angle between the axis and slant of my latitude. However, these would be at a 60 degree angle to the upper plate here, which I feel may weaken it. However, if I use ellipses, both bearings can be perpendicular to the upper plate, which should be stronger and easier to build. These ellipses would have to be perpendicular to the slant of the cone for this to be true.
I know that a cone where the axis is at 30 degrees to the slant should have an equation of 3(x^2 + y^2) = (\sqrt3 - z)^2. I know very little about 3D geometry, so I'm not entirely sure of this. This is what I got from Wolfram Alpha. Because I'm at 30 degrees north, presumably if I slice this cone with a plane with normal vector <0, -1, \sqrt3>, this will result in an ellipse perpendicular to the slant of the cone. (Again, I'm not entirely sure on this.)
I tried just setting the plane -y + \sqrt3 z + 1 = 0 equal to the cone to find the intersection, but I got an equation that isn't an ellipse. Is this the proper way to go about calculating the equations of these ellipses, or should I do something else?
Thanks!
So, I'm designing an equatorial platform mount for my telescope at the moment. I'm also going to use it for another telescope that I'm in the process of building.
I know that for both of the bearings, I can use small sections of two circles cut from a cone with an angle between the axis and slant of my latitude. However, these would be at a 60 degree angle to the upper plate here, which I feel may weaken it. However, if I use ellipses, both bearings can be perpendicular to the upper plate, which should be stronger and easier to build. These ellipses would have to be perpendicular to the slant of the cone for this to be true.
I know that a cone where the axis is at 30 degrees to the slant should have an equation of 3(x^2 + y^2) = (\sqrt3 - z)^2. I know very little about 3D geometry, so I'm not entirely sure of this. This is what I got from Wolfram Alpha. Because I'm at 30 degrees north, presumably if I slice this cone with a plane with normal vector <0, -1, \sqrt3>, this will result in an ellipse perpendicular to the slant of the cone. (Again, I'm not entirely sure on this.)
I tried just setting the plane -y + \sqrt3 z + 1 = 0 equal to the cone to find the intersection, but I got an equation that isn't an ellipse. Is this the proper way to go about calculating the equations of these ellipses, or should I do something else?
Thanks!