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Normal Vector for a cylinder

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    An infinite cylinder with radius 1 is aligned with axis Y, A point light source with intensity 1 is located at (2,2,10) An observer is located at (3,1,4). He is looking in the direction of the origin of the coordinate system. What are the coordinates of the point on the surface of the cylinder which the observer is looking at? Calculate the illumination at this point if the ambient are 0.1 0.65 0.3 and the specular exponent is 2 We also assume the ambient light intensity is 0.5

    2. Relevant equations

    First We find the intersection point. The line cast from the observer to the origin can be written parametrically as [x y z] = [3 1 4] + t( [0 0 0] - [3 1 4] ) = [3 1 4]( 1-t )

    The cylinder can be defined implicity x^2 + z^2 -1 =0
    Substituting the line equation into cylinder equation gives 25(1-t)^2-1=0 which has two root 4/5, 6/5. We choose the smaller root t = 4/5, which corresponds to the intersection point we want (x y z) = (3,1,4)*(1-t) =(3,1,4)*(1-4/5)


    actually this is copy out from my lecture note, I don't quite understand how to get this equition to find normal vector.


    3. The attempt at a solution

    Basically this is not a homework, is just an sample question from my lecture notes, preparing for exam and I don't quite understand certain part so could any one please help me clear the doubts thanks.
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2

    HallsofIvy

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    You say a light source is at (3, 1, 4) and immediately say "Heis looking in the direction of the origin". Where is the obvserver? Are we to assume that the observer is at the light source?
     
  4. Apr 21, 2012 #3
    Thanks for pointed out the mistake, I had corrected the mistake already. Thanks again.
     
  5. Apr 21, 2012 #4

    HallsofIvy

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    Assuming that, by "axis Y", you mean that the y-axis is the axis of the cylinder, then any point on the cylinder can be written as [itex](x, y, z)= (cos(\theta), y, sin(\theta))[/itex] where [itex]\theta[/itex] is the angle a line from the point perpendicular to the y-axis make with the x-axis. We can write that as the vector equation [itex]\vec{r}(y, \theta)= cos(\theta)\vec{i}+ y\vec{j}+ sin(\theta)\vec{k}[/itex].

    The unit normal vector, at any point, will be the unit vector, parallel to the xz-plane, [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{k}[/itex].
     
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