# Direction a marble would travel on a plane

• May 21st 2009, 12:06 AM
epiekarc
Direction a marble would travel on a plane
I am trying to calculate the direction a marble would travle if placed on a tilted plane. I need a formula to calcualte this with variables b, c and d. d is defined as an angle measured from a base line created by starting at point 0,0,0 to the perpendicular intersection of planes x and y.

I will always know what b and c equals, we are solve for d.

Here are some additional details

starting on a level plane (x) i have another plane (y) intersecting at b degrees.

From plane y i have another plane (z) rotated at 90 degrees and tilted c degrees.

I need a formula to calculate the direction the marble would roll if placed on plane z. the answer needs to be an angle (of direction of travel) on plane x.

Please see the attached image for claification. email me for clarification

epiekarc at yahoo dot com

Thanks for all the help,

Eric
• May 21st 2009, 06:05 AM
earboth
Quote:

Originally Posted by epiekarc
I am trying to calculate the direction a marble would travle if placed on a tilted plane. I need a formula to calcualte this with variables b, c and d. d is defined as an angle measured from a base line created by starting at point 0,0,0 to the perpendicular intersection of planes x and y.

I will always know what b and c equals, we are solve for d.

Here are some additional details

starting on a level plane (x) i have another plane (y) intersecting at b degrees.

From plane y i have another plane (z) rotated at 90 degrees and tilted c degrees.

I need a formula to calculate the direction the marble would roll if placed on plane z. the answer needs to be an angle (of direction of travel) on plane x.

Please see the attached image for claification. email me for clarification

epiekarc at yahoo dot com

Thanks for all the help,

Eric

1. This is only an attempt and I'm quite sure that I've forgotten to take something important into account.

2. There are three forces which act on the marble:
a) the weight: $w = m \cdot g$
b) a force acting parallel to the plane z: $|\vec z| = w \cdot \sin(c)$ ......... (dark blue)
c) a force acting parallel to the plane y: $|\vec y| = w \cdot \sin(b)$ ......... (dark green)

3. As far as I understand the question you are interested in the angle between the orange base line B and the initial direction of the movement of the marble projected into the plane x(?). If so:

$\vec z \perp \vec y$

That means the vectors $\vec z$ and $\vec y$ form a rectangle parallel to the plane z. If this rectangle is projected into the X-plane (nice name for a mathhelpforum :D) it becomes the rectangle which has the side length z' (light blue) and y' (light green) with:

$z'=w \cdot \sin(c) \cdot \cos(c)$

$y'=w \cdot \sin(b) \cdot \cos(b)$

4. The angle $\theta$ can be calculated by:
$
\tan(\theta)=\dfrac{w \cdot \sin(c) \cdot \cos(c)}{w \cdot \sin(b) \cdot \cos(b)} = \dfrac{ \sin(c) \cdot \cos(c)}{\sin(b) \cdot \cos(b)}$
• May 21st 2009, 09:44 AM
epiekarc
Not sure we have it.
I understand how z and y were drawn in, what i dont understand is how or why we can calculate the angle. I was thinking about this problem some more and thought that if we drew in a circle and used the intersection of z and y as the mid point then i should be able to calculate for the lowest z-axis point value of the circle. I just don't know how to figure that out.
• May 21st 2009, 11:29 AM
earboth
Quote:

Originally Posted by epiekarc
I understand how z and y were drawn in, what i dont understand is how or why we can calculate the angle. I was thinking about this problem some more and thought that if we drew in a circle and used the intersection of z and y as the mid point then i should be able to calculate for the lowest z-axis point value of the circle. I just don't know how to figure that out.

According to my previous post you'll get a rectangle in the X-plane where z' is parallel to the line of intersection between the plane X and the plane Y.

The angle d has the diagonal and the side y' as it's legs. Use the right triangle painted light grey to calculate the angle d:

$\tan(d)=\dfrac{z'}{y'}~\implies~ d= \arctan\left(\dfrac{z'}{y'} \right)$
• May 22nd 2009, 02:05 PM
epiekarc
Calc Check
I have some concerns about the assumptions. I am especially not sure that the angle in the z plane is the same as that in the x plane. It seems to me that as the z plane moves, for instance if angle b and c were both 5° or both 40° the angle of d would be 45° however in the x plane the intersection of z’ and y’ would become more obtuse as the z plane approached vertical. At vertical d becomes 180, or a straigh line.
• May 22nd 2009, 03:22 PM
epiekarc
A little more help
Ok, after some further testing I do believe the formulas you gave me were correct. One last question, at what angle does the marble travle down? In otherwords, at what angle does that line the marble travles down intersect the x plane?

Thanks for all the help.