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Hip Ridge Angle Continued

Hip Ridge Angle Continued

Hip Ridge Angle Continued

(OP)
Previously, I asked the question of how to figure the angle of a hip ridge.  I got several answers, some with working equations.  However, one MVP (who is not around any more I guess) gave two equations, one that worked and one that did not.  The one that did not work is the one I have been trying to figure out.

  cos(Theta) = (sin(A)*sin(B)*sin(C))+(cos(A)*cos(B))
    
If A = B, and C = 90  ( hip with identical slopes at a 90 degree corner) then no matter what the slope, I always get the same answer.  This seem to cancel out internally.  I have worked on this in my spare time to resolve this problem.  But this has proven to be an exercise in futility.
    Also, in the explanation of the equation Kapitan stated:
 
Roof (1) with gutter in X direction = 35 degrees = Angle (A)
Roof (2) with gutter in Y direction = 60 degrees = Angle (B)
Therefore Normal (1) has angles of 90,55 & 35 to X,Y & Z
      and Normal (2) has angles of 30,90 & 60 to X,Y & Z

The Cosines of these angles will values of i,j & k in each case;

   Normal (1) : i1=0.000000 , j1=0.573576 , k1=0.819152
   Normal (2) : i2=0.866025 , j2=0.000000 , k2=0.500000

Putting these values into equation to find Theta (Hip Ridge Angle)

      cos(Theta) = [(i1*i2)+(j1*j2)+(k1*k2)]

                 = [(0)+(0)+(0.409576)]

           Theta = 65.82 degrees or 114.18 degrees included angle.

       ( Theta is the angle between the two Normals
      so 180 - Theta is the angle between the planes )

In the case of the window bays being at 135 degrees then the Roof (2) is rotated about Z by 45 degrees = Angle (C)
The Direction Cosines can be rotated directly by;

              i3 = i2 * cos(45) = 0.612372
              j3 = j2 * sin(45) = 0.612372             
              k3 = k2           = 0.500000
But please notice that j2 = 0.  How did you arrive at 0.612372?

The thread that I am refering to is:
www.eng-tips.com/gviewthread.cfm/lev2/21/lev3/68/pid/507/qid/79917

I did get one response from LUTFI.  If you are still there can you help me.

  

RE: Hip Ridge Angle Continued

I should have been clearer in explaining what happens in the rotation of one window out of the orthogonal case, instead of just giving the result.

This is a simple rotation of one line (the Normal to Roof2) about a vertical element. These direction cosines can be transformed as though they are points, so the rotation of point X1 about an origin by an angle C to a new point X2 is;

                        x2 = (x1 * cos C) - (y1 * sin C)

                        y2 = (x1 * sin C) + (y1 * cos C)

Substituting i, j & k's for x, y & z's gives;

                        i3 = (cos 30 * cos 45) - (0.0 * sin 45) = 0.612372

                        j3 = (cos 30 * sin 45) + (0.0 * cos 45) = 0.612372

                       k3 = k2     

I hope this explains it. When doing any test case like this I would avoid an angle of 45 to base any calculations on as SIN 45 equals COS 45 which can be misleading.

This has been worked out by hand and then in BASIC code which not only prints out the results but links directly to a 3D model in CATIA V5.
Doing this proves each step as the rotated line becomes an element in the model, which can be analysed.

For all bay angles use the following equation ;

           CosTheta = (i1*i3) + (j1*j3) * (k1*k3)

Where
            Theta = Hip Ridge Angle
            A = Angle of Roof 1
            B = Angle of Roof 2
            C = Angle of Rotation from 90 degree corner.

If working out the 90 degree corner case then C = 0 degrees.   

            i1 = 0.0
            j1 = sin A        
            k1 = cos A

            i2 = sin B
            j2 = 0.0
            k2 = cos B

So, from above Rotation explanation,
 
            i3 = (i2 * cos C) - (j2 * sin C)
            j3 = (j2 * sin C) + (j2 * cos C)
            k3 = k2









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