Working out a point / segment on a sphere

  • Thread starter Thread starter rede96
  • Start date Start date
  • Tags Tags
    Point Sphere
Click For Summary
The discussion revolves around determining the optimal distance from the center line for a segment line on a rotating sphere, ensuring that intersection points a1 and a2 remain within a shaded area for specific percentages of the sphere's rotation. Initially, when the axis is at 0 degrees, a1 and a2 are always in the shaded area. However, when the axis is tilted to 60 degrees, the challenge is to find the distance x that allows a1 and a2 to remain in the shaded area 75% of the time. Participants suggest visualizing the rotation path and the shaded region to establish the necessary cutoff lines for the desired proportions. The conversation highlights the complexity of geometric relationships on a sphere and the use of trigonometry to solve the problem.
rede96
Messages
663
Reaction score
16
Hi, I was hoping someone could help me figure out the problem below. It is a bit of a long winded questions so please bare with me!

If you look at Fig 1 below, I have a sphere that spins about an axis in a clockwise direction. (the direction of the spin doesn't really matter) In this case the axis is pointing directly upwards, so the angle is 0 degrees.

You will notice the sphere also has a shaded area which covers the top half of the sphere.

If a put a 'segment line' perpendicular to the axis, which is at a distance x from the centre line, it will intersect the surface of the sphere at 2 points which I have labelled a1 and a2.

I know that as long as I place the segment line above the centre line then points a1 and a2 will always be in the shaded area as the sphere rotates.
fig_1.jpg

Now if I turn the axis by 90 degrees as in Fig 2 below, I can put a segment line at any distance x from the centre line, perpendicular to the axis, and points a1 and a2 will spend 50% of the time in the shaded area and 50% of the time in the unshaded area, as the shaded area doesn't rotate with the axis.
fig_2.jpg

So, I now turn the axis to 60 degrees as in Fig 3 below and I want to know at what distance x would the segment line need to be away from the centre line so points a1 and a2 would always be in the shaded area as the sphere turns.

I can do a bit of trig here and make a right angled triangle between the centre point of the circle, point a2 and the point where the segment line intersects the axis. I know the angle is 30 degrees and I know the radius r, so Cos 30 r = x

fig_3.jpg


In Fig 4 below the axis is again at 60 degrees. My question is at what distance x from the centre line would I need to place the segment line so points a1 and a2 would spend 75% of the time in the shaded area and 25% of the time in the unshaded area as the sphere rotates. I just can't figure this out!

Any help would be much appreciated.

fig_4.jpg
 
Last edited:
Mathematics news on Phys.org
I would suggest starting with a picture of the circle a1 and a2 trace out in their rotation. Consider where the shaded region will fall depending on your variable, then find the cutoff line which divides the circle into parts of 1/4:3/4.
 
rede96 said:
Hi, I was hoping someone could help me figure out the problem below. It is a bit of a long winded questions so please bare with me!

Sorry, I never disrobe with strangers.
 
LCKurtz said:
Sorry, I never disrobe with strangers.
haha :) Don't blame you!

RUber said:
would suggest starting with a picture of the circle a1 and a2 trace out in their rotation. Consider where the shaded region will fall depending on your variable, then find the cutoff line which divides the circle into parts of 1/4:3/4.

Thanks for the reply but to be honest I’m not sure how that will solve it?

On the sphere, where the line a1,a2 crosses the diameter of the shaded area, (the line that runs horizontally across the middle of the circle) let’s call that point a3.

I know that I can split any circle into 1/4:3/4 easily enough, it is just 90 degrees and 270 degrees along the circumference. This would be equivalent to the path a1 and a2 take. I also know that if I draw a line across any of the quadrants of that circle, the mid-point of that line that must intersects point a3 on the sphere.

But what I have no way of knowing (Or haven’t figured it out yet!) is how big the diameter a1,a2 must be so when that circle a1, a2 sits in the sphere it fits in the right place for a1 and a2 to travel around the sphere in the shaded area 75% and non shaded 25%

Anyway, up for work early, so will try and give it some more through during the week.
 

Attachments

  • upload_2015-3-15_21-36-1.png
    upload_2015-3-15_21-36-1.png
    220 bytes · Views: 588
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K