Geometry - Findng and marking a line through center of sphere

j_Bryan
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Beleive it or not i can not find anywhere on the internet how to do the following.  
I have a pool ball that is 2 1/4" in diameter. (This is an actual ball, not a picture or drawing), with this ball i would like to mark exactly a north and south pole, then scribe a line around the ball connecting the two points. Then would like to do this for the east and west poles. So when i am finished i have a ball with a cross if looking at from any angle. I thought i was really close when using a compass, in which i think this would be the correct route to take.

Please any suggestions would be great.
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If you use a string to measure the circumference you will be able to mark the poles using the ends of the string as the one pole and the exact middle of the string as the other.  The quarter mark of the string will mark the equator.
If you want to get extremely precise I'm sure we can find an answer for that also.
ozo
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Commented:
stretch a piece of string around the sphere, about 1/3 to 1/2 way around, mark the path of the string,
shift the string along the path and stretch it out again to extend the path, until you get all the way around.
mark a point along that path, and find a point at right angles to the path.
stretch a string between those points, and extend that second path all the way around
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Commented:
String?, i would like to be a litte more precise than a string, i have lasers which will project a vertical line on the ball, would just need to determine exact polar opposites.
Make a mark at any point. (The N pole)
Take a caliper (with inward pointing points) and set it so that the ball just goes between them. Put one point on the N pole.  Where the other point scratches is the S pole. Draw Your first circle.
The E pole is the point which is equidistant from all parts of that line. One way is to take a string the length of the distance between the N and S poles, and mark on it the halfway point. Put the two ends on various parts of the first circle. Mark where the halfway point lands. An average of these points will give you your E pole. Find the W pole as you found the S pole.

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Commented:
Aburr, getting closer to a solution i like...i will work with this for now, but i have a hard time beleiving that this can't be dont with a compass and somehow using the "small circle" and great circle principals.
Here is a trick to find the centroid of any object (I'll give you the whole process even though you'll only need the first part).

Attach a string (yes, a string again...stick with me though...to be very precise you could insert a screw with a hook, but that may ruin the ball).  That point is one pole.  Now take an ink pad or something that will leave a mark and slowly lower the ball over it until it just touches the ball.  Now that is the other pole...you now have both poles precisely.  Again, the exact half way mark gives the equator.

As far as the centroid is concerned, you just do the same thing again at any other point on the object.  Where the two strait lines intersect is the exact center of mass (the centroid).

Author

Commented:
Interesting approach.
My approach assumes the ball is of uniform density.
If you want three intersecting circles, one iach of x, y and z planes then there will be six equidistant points around the sphere where the lines cross: N, S, E , W and longitudes 90 and 270 at the equator. Imagine an inscibed cube with it's vertices at these points and you should be easily able to calculate the distances between them for marking with calipers or with a compass with folding legs.
iach="in each", blame my pda.
oops not cube, octohedron, but the principle is the same.

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Commented:
Imagine an inscibed cube with it's vertices at these points and you should be easily able to calculate the distances between them for marking with calipers or with a compass with folding legs.

uuhhh, sorry i have compass with folding legs, but the calculations are beyond me....if i knew where to start i could figure it out i'm sure..
notice how people glibly say 'you can easily' when really they are trying to avoid having to work it out themselves :7)
It has been a while, I found a page on wolfram that has loads of formula and one very useful fact, the regular octohedron, the one that fits inside your shpere, also fits neatly inside a cube with it's vertices in the centre of each face of the cube (I knew cubes came into it somewhere).
http://mathworld.wolfram.com/Octahedron.html 
So now imagine your sphere inside a cube and you want to mark the places where the sphere meets the centres of the faces of the cube.
The lengths of the sides of the cube are equal to the diameter of your pool ball (2 ¼ inch)... still playing for time here, I'm going looking for pencil and paper.
Distance between North pole and the equator is given by:
SQRT((diameter / 2)^2 * 2), which I make as 1.5909 for a diameter of 2 1/4
Please check this before you drill any holes.
 
Here is a rather rushed and scruffy diagram to show how I got that figure.
The triangle top right is formed from vertical lines prependicular to the cubes faces so the lengths are half the height of the cube which is the same as the diameter of the sphere. The hypotenuse represents the distance between the pole and the equator, or any point on that equator ie. where you want the centres of the other two circles.
Hypotenuse found by Pythagorus theorum is the square root of the sum of the squares of the other two sides.

pball.JPG

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Commented:
Robin D,

I have come to use a similar method as you have described except my figures are a little different.

Diameter of ball = 2.25
Circumfrence = 2.25 * pi = 7.068584
Cirumfrence / 4 for four equal points = 1.767146

when i transfered these numbers to my ball, using (yes a type of string), everyting seemed to matchup correct.

Once i got all my marks i then set my compass to the distance between two points, and you know if i remember correctly when i took the distance from the compass and measured it it was like 1 5/8"... or 1.625 which is pretty close to what you have..
I showed how to calculate the number s 'accurately, but knowing a number is correct two or three decimal points of an inch won't help you draw it any better. If you use your compass as a pair of dividers you will probably get a better physical result. Tech drawing at school even of a simple triangle usually ended up with one line being slightly out. There were limits to how close you could measure with a chipped plastic ruler.
Please don't take this as a condesending remark.  Just a thought:  to get a table level, you use a liquid level.  To find the exact half-way point on a piece of paper, you fold it in half (carefully).  The principle here is to just use the right tool at the right time.  Having worked on projects as an engineer, I can tell you that the carpenter's trick will often be more accurate than getting out a big tape measure an hoping you've got steady enough hands.  The math is sometimes the easy part...finding the best tool to get the best results is often times more difficult.

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Commented:
sl8rz,

I completely agree having the correct tool is essential, i have been going nuts trying to find the right tool...any suggestions.
How precise do you need to be?

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Commented:
is within a millimeter or two asking for to much....
thinking 'inside the box' again, and with a clever suggestion from number 1 son.
Draw a 90 degree cross on a piece of stout card. Using your compass, a compass cutter or a drill of the right size, cut a circle with its centre on the cross and the same diameter as the ball. Place the ball in the cutout and mark on the ball the four points where the lines of the cross meet the ball. Check the distances with the cranked leg compass to ensure they are the same and then mark top and bottom of the ball with small arcs at the same compass setting from all four points to set the position of the n and s poles. Use the same compass setting and three of these centres to mark your three x,y,z equators.

Commented:
Using card or some other suitable rigid material, create a cube whose internal measurements are 2.25" x 2.25" x 2.25".

Place the ball inside the cube so that each of the 6 internal faces of the cube touches the ball.

Working with each of the 6 outer faces of the cube, scribe two lines from corner-to-corner which intesect in the middle of the face.

Poke a sharp object, compass-point or marker, where the lines intersect on each of the 6 faces of the cube so as to make a mark on the ball.
 
I'd use either the string method I mentioned earlier (with a small drill hole you could really get precise).

Another way would be to find a corner of a book shelf or counter top where you can verify a 90 degree angle (do you have a square?).  Place the ball in that corner so that it is touching all three sides.  Now placing some inked paper in the corner between the ball and the side (under pressure this should leave a mark on the ball).  Now, holding the ball securely slide your square along the edge where the shelf or counter top meets the back wall (with the extending leg running headfirst toward the ball).  Slide it up till it almost touches the ball.  Ink the square where it will be touching the ball and sweep it back and forth until it just touches the ball.  That mark is one pole and the other paper will leave a mark for the other pole.  What's more, you'll also have two of your other crossing pole marks too if you ink where they touch as well.   I think this will get you within a couple millimeters.
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Commented:
Roll the ball down a smooth slope coated with ink and then saw through this line to obtain two hemispheres. Take the square edge of a piece of paper and place this on the edge of the flat circle and mark the two places where it crosses the circle and join these together (Thales' Theorem). Let hemisphere rest on the curved part and saw straight down through this line. Glue the four quarter spheres together. (I say use a saw but you could do a better job burning through it with your laser).
When you say you want to 'scribe' a line, if you mean other than with the pencil in your compass then you are going to want to hold the ball so it can revolve and have a marking tool held in a jig that can be postioned at the correct height.
The easiest way to hold the ball would probably be in a lathe if you have acces to one. Mount the ball between the headstock and a revolving tail centre using cup chucks and use a mounted tool post to hold the marking tool.
or bodge it like I just did...
I used a golf ball and tried holding this in a vertical drill stand using a golf tee in the chuck of the drill. I needed a way for the bottom of the ball to revolve so I tried a few options, middle of a roll of tape, another golf tee and the top off a 'sports' water bottle. Unfortunately none of these allowed the ball to revolve freely enough not to drift off centre whilst turning it. The option I tried that did work was using a revolving tail centre from my lathe, it is a very nice tool and cost me about £80 some 20 years ago so you won't want to buy one of these for a one off job. You might get it to work with a small ball bearing race perhaps from a car or motorcycle engine, but you will want it to be running fairly true ie. if its a really worn out one from an old engine it may not be good enough.
For the marker I simply held a pen in a small portable vice and positioned it by eye as I was only checking the holding of the ball. To set this up properly you will need to have the location marks on the ball (from the previous calculations) and ensure that you have them running 'on the equator' before setting the pen height to them. You could replace pen with a scriber or the cutting head of a dremel once you are confident enough to make cuts.
 
I set the ball up in this contraption purely by eye as I was only really checking for true revolution. It worked for me, but the slight non-squareness of the lines is noticeable. The pen holding arrangement coped with the lumpy surface of the ball. Sorry the pics are so dark but if I use flash it wont focus properly and I have temporarily lost my photo editing software.
Hope this is the arrangement of lines that you want.
 

equator.JPG
3rdLine.JPG
I brightened the picture a bit, you can now see the end of the drill chuck and the vice holding the marker pen a bit more clearly.
I didn't mention before and probably should, the ball was turned by hand and not under power. All you need is to revolve it carefully, spinning it with the drill when it is held like this is probably dangerous.
 

equatorretouch.jpg
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Commented:
Robin, how can you tell if the circle you've drawn is the great circle on the equator rather than a smaller circle like the tropic of cancer?

I can't think of anything better than string to mark a great circle on a sphere, although we can use good quality mathematical string I presume. A laser mounted perpendicularly from the sphere's surface would work as well, but how do you measure perpendicular? Once you've got one great circle the orthoganal ones are simple to construct using a pair of compasses (not a compass since a pool ball isn't magnetic) and more string.
Commented:
Okay.... I've been mucking about with a ball and compass. Here goes!

Diameter of ball = 2.25" therefore,



METHOD

Open your compass to a distance of 1.59099" (or as close to as the human eye can see....) (the Rule-Of-Thumb method may apply here).

1) Find any point on the surface of the sphere, called 'O1', draw a circle, C1' whose diameter is 2.25".

2) From any point on 'C1', called 'O2', draw a circle which intersects 'C1' and 'O3' and 'O4' and whose diameter is 2.25".

3) From either point 'O3' OR 'point 'O4', draw a circle which intersects ''C2' at 'O1' and 'O5' and whose diameter is 2.25".


    Point 'O1' = North
    Point 'O2' = West
    Point 'O3' = East
    Point 'O5' = South

Commented:
Oh, by the way, the quoted distance 1.59099" was found using Pythagoras' theory. ie,

Spere's diameter: D= 2.25"

Spere's radius: R=1.125"

Distance = SQR ( 2R^2 )

Commented:
Dur to a typo... I'm re-posting..



METHOD

Open your compass to a distance of 1.59099" (or as close to as the human eye can see....) (the Rule-Of-Thumb method may apply here).

1) Find any point on the surface of the sphere, called 'O1', draw a circle, C1' whose diameter is 2.25".

2) From any point on 'C1', called 'O2', draw a circle which intersects 'C1' at points 'O3' and 'O4' and whose diameter is 2.25".

3) From either point 'O3', draw a circle which intersects ''C2' at 'O1' and 'O5' and whose diameter is 2.25".


    Point 'O1' = North
    Point 'O2' = West
    Point 'O3' = East
    Point 'O5' = South

> how can you tell if the circle you've drawn is the great circle

Because you have already accurately marked the crosspoints using one of the above methods. You mount the ball, set the marker in its  tool holder to one equatorial mark and turn the ball on the drill's axis to make sure that all the points are on the same path. If they aren't then adjust the mounting of the ball until they are.
Once the ball is running true you can make your marks or cut grooves if you want.
This is an interesting question, the nearest I have seen to a solution is using an ornamental tturning lathe with an indexing system. Even those methods don't offer an easy way of finding opposing points on a sphere except for mounting it similarly to how I have and using the indexing facility to measure 180 degrees. Even then it won't guarantee that you are marking exactly on the equator. Compass and Pythagorus seems the easiest way. A bit like drawing those two dimensional 'flower petal' hexagons everybody draws with a compass. You can reassure yourself that it is correct when all the crosses meet in the same place.

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Commented:
t0t0

i am goig to try your method, as i think ths will be the most accurate, i will let everyone know shortly

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Commented:

1) Find any point on the surface of the sphere, called 'O1', draw a circle, C1' whose diameter is 2.25".

2) From any point on 'C1', called 'O2', draw a circle which intersects 'C1' at points 'O3' and 'O4' and whose diameter is 2.25".

3) From either point 'O3', draw a circle which intersects ''C2' at 'O1' and 'O5' and whose diameter is 2.25".



am i missing something here, in #2 you say to draw a cirlce which interect C1 at points 03 an 04, where are those points when in the one prevous step all we did is draw one circle so there would be only one point.
Just place your compass anywhere on the first equator you drew and another 'equator' of the same radius will cross the first one at two places.
From either of these cross points draw your third circle, at 90 degrees to both.
 
The accuracy of this method depends on your measurement of the diameter and your skill with a pair of compasses.
 

Commented:
j_Bryan

As RobinD states, drawing a circle from any point on 'C1' will intersect 'C1' at, and therefore create, two new points which are to be assumed 'C2' and 'C3'.

Unlike most others though (including RobinD) I avoided references to angles (degrees or radians) because such references are unnecessary.

Compasses vary in design. Some compasses are specifically designed to scribe lines onto spherical surfaces - their arms can be bent so that both the compass point and the scribe are at right-angles to the surface.

In my opinion, all that is required is a compass and the ball - no strings (how ludicrous!!), no rulers or folded paper tapes or anything else for that matter. As for a lathe... that really had me laughing.

I hope I have made myself a little clearer...


HERE ARE SOME ITEMS MENTIONED ABOVE WHICH MIGHT NOT BE USEFUL

String
Callipers
Screw
Hook
Inkpad
"Something"
Pencil
Paper
Drill
Chipped plastic ruler
Liquid level
Stout card
Compass cutter
Inked paper
Bookshelf
Square
Smooth slope coated with ink
Saw
Square edge of a piece of paper
Glue
Laser
Jig
Roll of tape
Golf tee
Sports water-bottle top
Small (unworn) ball bearing race taken from a car or motorcylcle engine
Pen
Small portable vice
Dremel

and finally, a means to lobotomise yourself !!
how many balls have successfully marked toto?

At least I test my ideas before I post them. If it makes you laugh then enjoy, but I didn't find your negative and snide comments at all amusing.

Commented:
RobinD

Sorry if you feel offended by my last comment - it was intended to be partly light-hearted and tongue-in-cheek.
Apology accepted. I would like to point out though that you say you make no references to angles, although you say you calculate the radii of your circles using Pythagorus' Thoerum which relies quite heavily on angles especially the 90° one in the centre. You are also relying on some circle theorums to assume that the two shorter sides of the triangle are equal (as I drew above).
Your suggestion that using string is ludicrous suggests that you don't really understand this problem or the difficulty in finding a great circle. The ideal answer would be a way to construct the circle and to be certain that it passes through both poles. Anything else is approximation, you still need to accurately measure the diameter of the ball before you can calculate the radius for your compass to draw the equator. Accurately measusing any sphere requires two flat planes that are absolutely parallel to each other so you can measure the distance in between them. Using calipers on a ball may not definitely be at the opposing points of the ball, and that is the core of the question.

Commented:
RobinD

>> "although you say you calculate the radii of your circles using Pythagorus' Thoerum" which relies quite heavily on angles especially the 90° one in the centre".

I disagree. The diameter of the ball is a known fact and therefore the radius is also a known fact. Furthermore, and as surprising at it might seem to you, the distance 1.59099" is also a known fact given that these facts are all known BEFORE we even pick up the ball and compass.

in my answer, I start by saying "Open your compass to a distance of 1.59099"...". There is no need to use string or anything to determine something we already know.

If the asker had not stated the diameter of the ball then I may have provided a more elaborate solution to the one I did.

>> "You are also relying on some circle theorums to assume that the two shorter sides of the triangle are equal (as I drew above)".

Sorry to disagree with you again. I didn't assume any such thing. As implied in the question, the ball is perfectly spherical. This is arrived at by way of logical reasoning - (1) because the asker did not state otherwise, and (2) it's a known fact pool balls, billiard balls and snooker balls are near-perfect spheres. Allow me to give you an example:

I would bet the accuracy and perfectness of a machined pool ball compared to the accuracy of your placing a golf ball between a golf tee in the chuck of a vertical drill and a revolving tail centre from a lathe while holding a marker in a vice is probably like comparing a walk to your local corner shop and a 9-year flight to Uranus. (Please take that as an approximation).

>> "Your suggestion that using string is ludicrous suggests that you don't really understand this problem or the difficulty in finding a great circle".

Nonsense! Of course I understand the problem. See my previous reply (re: the cube).

>> "The ideal answer would be a way to construct the circle and to be certain that it passes through both poles".

I do believe steps 1 - 3 (in my proposed solution above) would more than satisfy the requirements to an accuracy well below the stated 2mm.

>> "Anything else is approximation, you still need to accurately measure the diameter of the ball before you can calculate the radius for your compass to draw the equator".

No you don't! We already KNOW the diameter of the ball! Please re-read the question!

>> "Accurately measusing any sphere requires two flat planes that are absolutely parallel to each other so you can measure the distance in between them".

(Not if you're going to use a chipped plastic ruler) however, I do agree that is a good method for measuring the diameter of a sphere.

I also believe it's possible to find the size of a sphere by it's weight.

>> "Using calipers on a ball may not definitely be at the opposing points of the ball, and that is the core of the question".

What if the calipers had articulated flat surfaces and micrometer measuring scale?


t0t0, please meet me here to save spoiling this thread:
http://mobile.experts-exchange.com/Q_26320557.html

You may need to re-link to it as that takes you to the mobile interface.

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Commented:
Ok, everyone here is a few snapshots of the approach i used. Which was t0t0's, Thank you, and you have been awarded the points.  I used a compass and a charcaol pencil.
IMG00122-20100711-0550.jpg
IMG00123-20100711-0550.jpg
IMG00124-20100711-0551.jpg
IMG00125-20100711-0552.jpg
IMG00126-20100711-0552.jpg

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Commented:
Thanks
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Commented:
You could at least have cleaned your fingernails ;)

Commented:
j_Bryan

Thank you for applying my method and for posting an interesting question.

I enjoyed the debate as much as assising in finding a solution which has proved successful.

A sharper (thinner) line would have given a greater (or finer) degree of accuracy however, it's a great attempt given the tools available.



RobinD

The question is now closed. It was a pleasure debating the question with you.
>Nonsense! Of course I understand the problem. See my previous reply (re: the cube).
You can construct a perfect cube from cardboard that exactly fits the ball? The cube idea was taken from one of my previous observations. I didn't suggest it as a practical solution, partly because of the difficulty of actually building an exact  cube of these dimensions and partly because you are still relying on an accurate measurement of the diameter of the ball.
Your use of Pythagorus followed my suggestion and the diagram that I posted to explain how to apply it. Again it still relies on an accurate measurement of the ball, but I did do the math for you and calculated distance between the points on the ball to more decimal places than were needed.
If you are going to use other peoples work to assist your own answer then it would be polite to at least acknowledge the fact, and if you claim you didn't see my posts then you are admitting that you didn't read the thread.
Your answer was chosen as the most useful by the asker and I don't have a problem with that. What I do find very annoying and rude is your copying of other peoples work without acknowledgement and in the rudeness with which you commented unnecessarily on other peoples posts.
That wasn't a debate, a debate does not have one of the participants calling another's comment 'nonsense', I offered you the chance to meet somewhere outside this thread for a discussion, but you apparantly declined.
 

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