Projection of area onto a plane

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This problem refers specifically to http://books.google.com/books?id=W9...ss components on an arbitrary plane"&f=false".

The text comments that Area BOC = Projection of area ABC on the yz plane, and that this equals the area of ABC times the direction cosine in the x-direction of the normal vector to the plane, and so on for the other faces.

Similarly, http://books.google.com/books?id=Gl...ss components on an arbitrary plane"&f=false" states that "it can be seen" that the ratio of areas Ax/A = nx, etc.

I'm looking for a mathematical verification that this is true. Most likely because it is a math issue, the authors deemed the explanation unecessary in a solid mechanics text. My linear algebra background is pretty weak, so you may need to include some fundamentals in order for me to follow.

Thanks in advance.
 
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on Phys.org
One way to understand it is to look at the two dimensional case. Project a line segment in the xy plane onto the x (or y) axis and compare the length of the original segment to its projection. This is an exercise in elementary trigonometry. The problem you are asking about is an immediate 3-d analog.
 
I understand the length ratio of the projection of a line onto an axis is the direction cosine (cos of angle between line and the axis).

In fact for this case, the unit normal vector to the plane has a projected length nx along the x-axis.

I just don't see how that logic extends to these triangular areas. The value nx is the direction cosine between the unit normal vector and x-axis, whereas here we are projecting a planar area normal to that vector onto the yz plane.
 
Perhaps I might rephrase my previous post.

Wouldn't the immediate 3-d analog to the projection of a line segment in 2-d space onto an axis (essentially the dot product with a unit vector along the axis) simply be the dot product of a line in 3-d space (i.e. a vector in 3-d) onto another vector (say unit vector along one of the 3 axes) in 3-d?

Now I've done projections of lines onto lines, and lines onto planes. But I haven't found much literature on projecting plane areas onto other planes.

...

Incidentally I have found a way to prove that the ratio Ax/A = nx.

Let (see figure 1.6 in link 1):

V = volume of the tetrahedron OABC
A = area of face ABC
Ax = area of face OBC
OA = length of line from origin to vertex A
ON = length of line from origin to point of intersect on face ABC, traveling normal to the face ABC
a = angle between ON & OA

V = 1/3*A*ON = 1/3*Ax*OA
noting the volume formula for a tetrahedron

thus Ax/A = ON/OA = cos(a)
due to right triangle OAN

noting that this is the same angle that gives the direction cosine of the normal vector to the plane
cos(a) = nx = Ax/A


I arrive at the conclusion by simple visual inspection of the geometry and I would still like to see it demonstrated how an alternate solution can be related to the concept of "projection".
 
Hi ptd! :smile:

Imagine two planes, one horizontal.

They meet in a line. On the sloping plane, draw a rectangular grid of lines parallel to that line and perpendicular to it.

Drop perpendiculars from that grid onto the horizontal plane.

Obviously, that will also be a rectangular grid, and the "perpendicular" lines will be the same distance apart on both planes.

But the "parallel" lines will be closer, by a factor of the cosine of the angle between any pair of them (you know how to prove that! :wink:).

So the rectangular grid by which we measure areas on the horizontal plane (using both the intuitive definition of area and the rigorous definition) is squashed by that cosine factor in one direction only, and accordingly all areas are squashed by the same factor. :smile:
 
Thank you! clear as day now.
 

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