Okay, you put a negative cylinder through a positive sphere, duplicate it and rotate to make a new hole, then keep doing this until there are holes everywhere at 0, 45 & 90 degrees respectively.

But...how can I acurrately put a hole at that one spot? Off the top of my head I always thought oh, that should be 45 degrees over, 45 degrees up. But that isn't working, (at least by turning a cylinder from 0 to 45 dgrees, and then trying to take it up 45 degrees).

Any geomtry brains out there know how I can place this?

I just kept eye-balling it, until I got a cylinder at the following degrees;

(X)-55, (Y)45, (Z)0

That seems to visually hit it perfectly.

Attached Link: Spherical trig.

..... but your way looks easier ;)

Part of my brain crawled screaming outta my ear, just from looking at that page...;o) 

It had that same effect here, don't you worry... my forebrain went into reverse so hard I have an extra bulge on the back of my head now.

roflmao.....

I think part of the problem is the way Bryce handles rotations through 3-axes... it gets really weird as it tries to simplify things.

Yeah, I was going from 0 to 45 degrees, and then trying to take the opposite side 45 degrees.

Taking one side up 45 and then just spinning it was what got me (almost) where I needed to go.

More than one way to skin a cat. (lol...no offense Mahray)

Couldn't you just rotate the sphere 45 degrees to make the line your arrow indicate horizontal and then copy and rotate your current holes 45 degrees.

Yet, I did eventually get the cylinder at 45 & 45 degrees (where I thought it would go), but it never lined up with the crossing of the lines, which was the real confusing part, on my part.

For some reason that X degree needed to be at 55 instead of 45. Once I got that in place, mirroring the other holes was no problem at all.

Finished now. I also went ahead and placed a Tori on each side, of each band to round off their edges (whew!)

....and very cool it looks too. Just shows that line of sight can be better than maths!!!! :LOL:

As long as it looks the way I want.  :o)

Looks like your new hole anted to be at two thirds of an arc, rather then at half... around 60 deg?  Or you can look at an equilateral triangle, and find it's orthocenter point.
Off the top of my head, I don't think the arc anle angle to the orthocenter is exactly 60 deg, but it may be close to it...  You may have been looking for 1 radian... 57-17-45  Center of mass may be at 60 deg.

http://en.wikipedia.org/wiki/Triangle
http://mathworld.wolfram.com/EquilateralTriangle.html
http://mathworld.wolfram.com/Orthocenter.html

If I wasn't comfortably curled up on the couch with my kities, I'd actually get up and figure his out exctly, but looks like you have an answer  :)

LOL, they didn't sendme to enhineeeering schfool fer nuttin'    :m_laugh:

Well, those bands cross right where the hole should be (post #2), and at 55 degrees, the cylinder nails it perfectly.

I don't know....perhaps if one could measure the distance between the 55 degrees hole and the other 3 holes around it, perhaps the distance would not match up, and it would be closer to one of the holes compared to the other?

Ah, at this point, it works, lol. Anything more "correct" and I would have to burn perfectly good brain cells, and them is reserved fer drinking, lol.

When I get to the offic tomorrow, I'll have to do it exactly to specs... I'm dying to know how close my guesstimate was to the real solution. I should know this stuff, I do it for work.   (just testing myself)

When you eyeball it, you probably couldn't tell  a difference between 55 or 57 degrees, so what you did is purrrrfect  :)

Maybe I should take up drinking instead.... hmmmm......

You got me wondering.....I'll do a test render of the difference between 55 or 57 degrees, be back in...say 15 minutes....;o)

After a set of  extremely close rendering and aligning it with the straight edge of my plop render window....the true degree seems to be very close to -54.725 degrees.

Go figure.  ;o)

ooh goody! that link with the spherical trigonometry reminds me of highschool mathematics in my final year! :biggrin: I have fond memories of that. One of the only subjects that I really really had to work hard on to even just barely pass it. I loved it, such a challenge! Now, only 3 years and 3 months later, I forgot most of it again though... The value is indeed not 55 degrees btw, and u should be able to reconstruct the exact coordinates with pythagoras' formula 'cuz I did that once before with the help of my dad. It's a brain teaser though, I cant remember how I did it... 54.725 is close, but not exact yet from what my memory tells me... but i bet it's close enough I think you can also just reconstruct it by grouping 3 cilinders with equal length along the x, y and z axis, their center located at the center of ur sphere. Then rotate the group 45 degrees around the z axis, ungroup it, regroup it (to reset the groups bounding box), and then rotating it 45 degrees over x, ungroup it again, regroup it again, and finally rotating it 45 degrees over the y axis. My brain says that should work... ooooooh and now that I typed that I just remembered how to calculate it! :biggrin: haha O'm too lazy though, it's kinda complicated :-P but think of it this way: U can construct a triangle between the center of the sphere, and the points where the sphere crosses on 2 axis. A line rotated 45 degrees over the 3rd axis running paralel to the triangle and trough the center of the sphere will also run through the center of the line of the triangle closest to the outside of the sphere. This is the same for all three triangles u can construct that way. (These lines should split each triangle in 2 straight edged triangles, thus u can use A^2+B^2=C^2 and the rules on using sine, cos and tangent to calculate the lengths of the sides of the triangle.) A 4th triangle can be constructed from the lines of the 3 triangles that are closest to the outside of the circle will be a triangle with each side at equal length, making each of this 4th triangles corners 60 degrees exactly. The earlier mentioned lines splitting the other 3 triangles in 2 straight edged triangles each will run through the center of the sides of this equal triangle also. Constructing lines through these centerpoints directly to the corner on the other side of the equal triangle will point out the center of the triangle. Using the same pythagoras formula and the use of sin/cos/tangent as before we can calculate the coordinates of this center point which allows us to construct a triangle with one line along an axis, one line from the center of the sphere to the center of the 4th triangle, and one line to conect the ends. You can then again use sin/cos and tan/ to construct the angles of all these lines, including the one goin from the center of the equal triangle and the center of the sphere. (which is the line along which u want ur cilinder placed) :biggrin:

Quote - Looks like your new hole anted to be at two thirds of an arc, rather then at half... around 60 deg?  Or you can look at an equilateral triangle, and find it's orthocenter point.
Off the top of my head, I don't think the arc anle angle to the orthocenter is exactly 60 deg, but it may be close to it...  You may have been looking for 1 radian... 57-17-45  Center of mass may be at 60 deg.

http://en.wikipedia.org/wiki/Triangle
http://mathworld.wolfram.com/EquilateralTriangle.html
http://mathworld.wolfram.com/Orthocenter.html

If I wasn't comfortably curled up on the couch with my kities, I'd actually get up and figure his out exctly, but looks like you have an answer  :)

LOL, they didn't sendme to enhineeeering schfool fer nuttin'    :m_laugh:

Ah! A girl with maths skills! :biggrin: If u were 17 I'd be in love

Hmmmmm, I feel like I'm in Eureka.....Looks good, either way you want to do it......so, what's it for?

I once had a dose of "Spherical Trig", but my doctor gave me a scrip for a course of pills, and after a week it went away... Still left a small hole where it shouldn't be, but my Doc said  "Aahh.. that's just Gymbal Lock - ignore it and after a few years you won't notice it"

Didn't understand any of it, and after a few years I still don't..

Cheers,
Diolma

so, what's it for?

Nothing really. I was just doodling with booleans, it's kind of my way of relaxing, taking a break from my work.

Quote - I was just doodling with booleans, it's kind of my way of relaxing, 

I think you need to get out of the house more. Take up Bowling. Interact with things that talk back.

Oh, trust me...my girlriend talks back enough, rofl...

Ah! A girl with maths skills! :biggrin: If u were 17 I'd be in love

You should hear me recite multiplication tables with a very slight foreign accent and an innocent look on my face.  A real turn-on....  not!

Agent, Sowwy, I got distracted at work today (believe it or not, with work), and didn't do the calc...

Can you recite the multiplication tables in a pair of paint streaked coveralls, gumboots and a balaclava? Using a husky Marilyn Monroe "Happy birthday Mr President" voice. Juggling CHAINSAWS.

I'll go you halves on the video sales.

Quote - Part of my brain crawled screaming outta my ear, just from looking at that page...;o) 

I'm sure I could actually feel the hairs on my head standing on end at one point after looking at that.  Must have been from the heat of my brain melting...

You know, I sometimes wonder if those sort of things actually mean anything or if people just post them simply to cause a load of brain melt.

(nods wisely)  Yeah, that's it.

Quote - You should hear me recite multiplication tables with a very slight foreign accent and an innocent look on my face.  A real turn-on....  not!

Oh, but I'm from the netherlands so I bet my accent is.. ahem 'bigger then yours' :tongue1:

Too bad I already have an affair with Angela hahaha :biggrin:

Btw, I just figured out something really stupid! While all this time i was breakin my head over trigonometry and stuff like that i forgot that it's just waaaay easier to use simple lineair algebra!

The center point of that equal triangle can be calculated so easily! All that's needed is to set up lineair formula's for the vectors of the three construction lines from my previous theory which should cross at the center of the equal triangle. Once u know the coordinates of the center point of the equal triangle you only have very little trigonometry left to do. And even then it's a simple rudimentary calculation. There's no real need for all the complicated sin and cosine formula's from the spherical trig website!

Strange how the simplest solutions are so often the last to come to mind...

Quote - > Quote - Part of my brain crawled screaming outta my ear, just from looking at that page...;o) 

I'm sure I could actually feel the hairs on my head standing on end at one point after looking at that.  Must have been from the heat of my brain melting...

You know, I sometimes wonder if those sort of things actually mean anything or if people just post them simply to cause a load of brain melt.

(nods wisely)  Yeah, that's it.

Imagine the horror seeing that on ur national highschool final exams