# Planar intersections

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• Jan 11th 2011, 08:09 AM
conorc
Planar intersections
Hello all, i wonder if you can help me with a problem im having.

I have a disc and three planes (representing a road, building and roof). For this problem, the disc can be considered a plane.

http://i3.photobucket.com/albums/y60/con20or/Fig1.png

I know the orientation of each of the planes. The disc is orientated 45horizontal x 45vertical. I can caluate surface normals for each element. With these I can calculate the dihedral angle between the disc and the road, and also the disc and the building, but what im really looking to find is when looking directly at a vertical surface (middle image) what angle does the plane make on the vertical surface? Its approx 35^ in this case.

This is really bugging me, i'm sure theres an easy way of doing this.

• Jan 11th 2011, 08:16 AM
Ackbeet
Quote:

but what im really looking to find is when looking directly at a vertical surface
Do you mean vertical view?

Quote:

what angle does the plane make on the vertical surface?
Do you mean what angle does the disc make with the vertical?
• Jan 11th 2011, 08:24 AM
conorc
Hi Adrian, thanks for getting back to me so quickly.

1. I suppose you could call it vertical view, im not sure, you might mean something else...the middle view is looking at it as though as you were standing up.

2. I have calculated the angle the disc makes with the vertical, it was one of the dihedral angles i mentioned. The problem comes when i try to calculate the angle the red disc makes ON the vertical surface.

Sorry, I know this isnt very clear. To try and clarify it a bit, the red disc intersects the vertical white surface at a number of points. I would like to find the slope of the line ( slope in relation to the horizontal as shown in the middle image) that these points make. I have no x,y,z coords at all, just the angle of each of the planes and the disc.
• Jan 11th 2011, 08:36 AM
Ackbeet
Quote:

vertical white surface
Ahh. That was the missing piece: knowing that the white surface was vertical. Your last paragraph was very clear, thank you. So, just to make sure we're all on the same page, let me restate the problem.

Given: a vertical plane, a horizontal plane, and one more plane (call it P) oriented at a 45 degree tilt from both the vertical and one horizontal.

Let $\displaystyle \ell_{1}$ be the line of intersection between the horizontal and vertical planes. Let $\displaystyle \ell_{2}$ be the intersection of P with the vertical plane.

Determine: the angle between $\displaystyle \ell_{1}$ and $\displaystyle \ell_{2}$ as measured in the vertical plane.

Is that correct?
• Jan 11th 2011, 08:46 AM
conorc
If $\displaystyle \ell_{1}$ is the point where horizontal, vertical and p all intersect, then yes, that sounds right.
• Jan 11th 2011, 09:02 AM
Ackbeet
The angle between a point and a line is undefined. The angle between two lines is defined. Three planes that have linearly independent normal vectors will meet at a point. So my line $\displaystyle \ell_{1}$, in order to be a line, cannot be where all three planes intersect, but only where two planes intersect.

Make sense?

So, you say you've computed normal vectors to all the planes. Could you please type those out for me?
• Jan 11th 2011, 09:19 AM
conorc
Sorry, took me a minute to sketch it up, that is what i was trying to say when i included l1.

http://i3.photobucket.com/albums/y60...or/image-1.png

$\displaystyle \ell_{1}$ is the line of intersection between the horizontal and vertical. I've made it a point here, it doesnt have to be, but for my angular measurement i have to measure from the point of intersection of the horizontal plane, the vertical plane and the disc up to t to $\displaystyle \ell_{2}$ which is the point of intersection of the disc and the vertical surface.

Its the angle 'a', formed at l2-l1-x.

I suppose i dont need the point to be there exactly, but i hope you see what im getting at...clear as mud maybe.

The surface normals i have are

For the ground
n_x = 0
n_y = 0
n_z = 1

For the wall
n_x = 1
n_y = 0
n_z = 0

and for the disc
n_x = 0.5
n_y = 0.5
n_z = 0.7071
• Jan 11th 2011, 09:35 AM
Ackbeet
Ok, I understand what you're getting at. The disc $\displaystyle n_{z}$ is, I presume, $\displaystyle \sqrt{2}/2,$ right?

The equation of the vertical plane is then $\displaystyle x=0,$ and the horizontal plane is described by $\displaystyle z=0.$ We will take the origin to be your L1, such that the point of intersection of all three planes is the origin. Then the equation of the plane is $\displaystyle \hat{n}\cdot\mathbf{x}=0,$ or

$\displaystyle \displaystyle\frac{x}{2}+\frac{y}{2}+\frac{\sqrt{2 }\,z}{2}=0.$

Now then, the equation of L2 we can obtain simply by setting $\displaystyle x=0$ in P's equation thus:

$\displaystyle \displaystyle\frac{y}{2}+\frac{\sqrt{2}\,z}{2}=0,$

or $\displaystyle y+\sqrt{2}\,z=0,$ or

$\displaystyle z=-\dfrac{y}{\sqrt{2}}.$

Since we're taking the angle between this line and the line x, we just want the angle whose tangent is equal to the slope of this line. That is, we solve

$\displaystyle \tan(\alpha)=-\dfrac{1}{\sqrt{2}},$ which implies that

$\displaystyle \alpha=\tan^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)\approx -35.2644^{\circ}.$

The slight discrepancy between this number and the number on your drawing might be chalked up to inaccuracy in the drawing, perhaps? It's pretty close.

Does this make sense?
• Jan 12th 2011, 02:08 AM
conorc
Yes, thats exactly it! I only took a rough measurement. Thanks for your help. Im a bit rusty on all this, so to make sure I understand the process, I have some questions if you dont mind?

1. $\displaystyle \sqrt{2}/2$ is 0.7071, but i found 0.7071 through multiplying rotation matrices in MATLAB. You seemed to be familiar with $\displaystyle \sqrt{2}/2$, is that a common number in geometry?

2. You mention the vertical as $\displaystyle x=0$ and horizontal plane as $\displaystyle y=0$, but in each of these cases, the normals were actually the opposite, i.e. n_x = 1 and n_y = 1. Was this a typo or have you substituted the normals into the ax + by + cz =0 formula, which when worked out give you the $\displaystyle x=0$ and $\displaystyle y=0$?

3.$\displaystyle \hat{n}\cdot\mathbf{x}=0$ is this a general formula? And p then becomes x? Or is it the $\displaystyle x=0$ of the vertical plane.

4. How does this $\displaystyle z=-\dfrac{y}{\sqrt{2}}$ become this $\displaystyle \tan(\alpha)=-\dfrac{1}{\sqrt{2}}$?

• Jan 12th 2011, 03:41 AM
Ackbeet
1. It's a very common number. $\displaystyle \sin(\pi/4)=\sqrt{2}/2.$ That's probably where your rotation matrices got the number, since you definitely have 45 degrees floating around in this problem.

2. Not a typo. You have to distinguish between the equation for a plane, and the normal vector to that plane. If I have a normal vector $\displaystyle \hat{n}$ to a plane, and a point $\displaystyle \mathbf{x}_{0}$ on the plane, then the equation of the plane is given by $\displaystyle \hat{n}\cdot(\mathbf{x}-\mathbf{x}_{0})=0.$ Basically, this equation says that the points in the plane are all those points where, if you took the vector from $\displaystyle \mathbf{x}_{0}$ to that point, it's perpendicular to the normal vector. In our case, $\displaystyle \mathbf{x}_{0}=0,$ since I put the origin in the plane. Make sense?

3. I think my answer to 2 may have answered this question also, but let me know if you have more questions.

4. Ok. You've got the equation of a line: $\displaystyle z=-y/\sqrt{2}.$ Compare to the general equation of a line: $\displaystyle z=my+b.$ Evidently, $\displaystyle m=-1/\sqrt{2}$ and $\displaystyle b=0,$ right? So the slope of the line of intersection is $\displaystyle -1/\sqrt{2}.$ The slope is the rise over the run. If you imagine a right triangle with rise of $\displaystyle -1$ and run of $\displaystyle \sqrt{2}$ (so it'll be pointing downwards), the angle $\displaystyle \alpha$ will be the angle opposite the rise. But the rise over the run will just be the tangent (opposite over adjacent is exactly rise over run here) of alpha. So that's how I got that equation.

Make sense?
• Jan 12th 2011, 05:31 AM
conorc
Perfectly, you're very good at explaining things.

So if the wall was angled slightly, either rotated in horizontal or tilted in vertical, could it be solved? You would not be able to substitute 0 in that case which would leave you with an extra variable...Or could you simulate a wall movement by changing the normals of p..?
• Jan 12th 2011, 05:39 AM
Ackbeet
Yes, it could be solved in that case. I think that instead of plugging in x = 0, you'd have a system of equations to solve (the equation of the wall would be more complicated than x = 0), but that's pretty routine.

Any other questions?
• Jan 12th 2011, 05:47 AM
conorc
Could you expand on what you just said about the system of equations?? Do you mean simultaneous equations? For example, if the wall had normals of

n_x = 0.5
n_y = 0.5
n_z = 0

(Im not sure if they are valid, im trying to describe a wall at 45^ to the original)
• Jan 12th 2011, 05:58 AM
conorc
Quote:

Originally Posted by conorc
Could you expand on what you just said about the system of equations?? Do you mean simultaneous equations? For example, if the wall had normals of

n_x = 0.5
n_y = 0.5
n_z = 0

(Im not sure if they are valid, im trying to describe a wall at 45^ to the original)

it just occurred to me that if the wall was angles 45^, the angle i was looking for would also be 45^ so that wasnt a great example. its the procedure im interested in.
• Jan 12th 2011, 06:16 AM
Ackbeet
And, it just occurred to me that more has changed than the wall's angle. With the wall tilted, how do you want to measure the angle that the intersection of the disc with the wall makes with the intersection of the wall and the ground? Are you still taking the angle as the projection onto the vertical (that is, where the wall was when it was vertical)? Or are you measuring the angle along the now-tilted wall?
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