how do i figure the angle of a ridge cap along a hip ridge

huangyhg

how do i figure the angle of a ridge cap along a hip ridge
does anyone know what combinations of formulas i need to use to figure the angle for a bent plate to cap a hip ridge? currently i have been using acad to draw in the roof planes and then do some measuring to get the angle.  but i am looking for something a little easier, and of course something cool that works for any combination of roof pitches.  i also need this infomartion so i can figure the structural property of the angle once it is bent.
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well, manually, here's how i would do it (its hard to describe so i'll go step by step):
1.  looking down in plan view, draw your ridge line at its proper plan angle and indicate on your drawing the slope directions.
2.  place a dot on the ridge line - point a.
3.  draw a line, perpendicular to the ridge line, through point a.  call this line ll.
4.  from point a, draw a short 1' line parallel with the sloped roof on one side of the ridge - call this line l1.  place a dot at the end of line l1.  call this point b.  point b will have the same elevation as point a.
5.  do the same for the other side - line l2 and point c.
6.  from point b, draw a line parallel with the slope, down the roof slope, to a point that intersects line ll.  draw a dot at the intersection.  this is point d.
7.   do the same on the other side of the ridge line for point e.
8.   calculate the distances of the lines and then calculate the elevations of each point d and e.
9.   from this you have the distances from point a to point d and from point a to point e.  you also have the elevations of a, d and e.  from this you can calculate the angle from vertical on each side.
all this could be put together in a spreadsheet to use again in the future.
just to clarify:
1.  are you addressing a hip ridge?  i think you are, i just wanted to make sure.
2.  i got lost on item 4. "draw a short 1' line parallel with the sloped roof on one side of the ridge".  after this line, later you say that eventually i will need to draw a line that intersects ll.  i am confused.  i know...
never mind.  i found an awsome site
when i say "parallel with the sloped roof" i mean to draw a line from point a along the roof such that this line stays at the same elevation - just like a contour line along your sloped roof.
if a and b are the angles (in degrees) of the two roof planes to the horizontal and theta is the hip ridge cap included angle then
                theta = arccos((-cos a) * (cos(b))
i eventually found the website that rsteel13 mentioned, it appears to be,
  
here is a very simple and accurate (to the minute) method to determine angles:
you will need: 1. horizontal distance of the slope and 2. height of the slope. 3. a calculator (i use a ti-30xa scientific).
example: lets say we have horizontal distance of 24' and a total height of 6' (assume this is a hip roof, run and height, but can be used for any slope).
divide the height 6' by the run 24' = 0.25 (tangent)
hit (inv) (tan) = 14.03 degrees (this is to a vertical element, add 90 degrees to get the angle to a horizontal element).
14.03 + 90 = 104.03 degrees
to convert to degrees and minutes multiply 0.03 x 60 =1.8
thus the angle is 104 degrees, 2 minutes.
print and save this reference, it has served me well.
oops, correction:
in my last post, i said " this is to a vertical element, add 90 degrees to get the angle to a horizontal element" this is not correct. subtract from 180 degrees, this gives the proper angle.
thus: 14 degrees 2 minutes - 180 degrees = 165 degrees 58 minutes.
kapitan,
  can you please explain further the variables i,j, & k?
for example a bay window where the walls are on a 45^ angle from each other.
cos(theta) = ((i1*i2)+(j1*j2)+(k1*k2))
i really liked your simplified formula. i have tried it several times and works very well.  but now you have me interested in the more involved equation.
to solve the problem for the case you have suggested means that the directions of the normals to the two planes have to be found, so i will try and give some background.
for any any line or vector in space there are three angles between itself and the axis system that it is in. the term direction cosines refers to the cosines of each of these angles. here i call them i,j,k but they may also be called l,m,n. it doesn't matter what they are called, the important thing about them is that the sum of their squares is 1. the cosines of the angles between the line and the axes x,y,z are i,j,k respectively.
to find the direction cosines of a line that starts at (x1,y1,z1) and finishes at (x2,y2,z2) then the distance (d) between the two points is divided by itself to become a unit length (1). this means that the orthogonal virtual 'box' that encloses the line has dimensions, and if these are divided by d then these are equal to the direction cosines.  
so, if a line of length 1 (inches, metres, etc.) goes from one point to another in space and the values of the start point are considered to be (x1,y1,z1) and the finish point (x2,y2,z2) then the length of the line is;
                        d = [(x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2]^0.5

then the direction cosines (i,j,k) are;
                        i = (x1-x2)/d
                        j = (y1-y2)/d
                        k = (z1-z2)/d
and
           i^2 + j^2 + k" = 1
in the original thread the two roof slopes were at rightangles to each other - the gutter of one being along x of a local axis system and the other along y. therefore the normals to the roof planes both lay on two of principal planes (xz and yz) which means that in each case one of the angles is 90 degree, the cosine of which = 0.  
the origin of the axis system is at the lower end of the hip, which is also the intersection of the gutters.
giving some values to the roof slope angles will make it easier to see what is happening;
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 bewteen 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
note : the sum of each of these values squared is 1
putting these values into equation gives;
# using (i1,j1,k1)&(i3,j3,k3)
      cos(theta) = [(0)+(0.351242)+(0.409576)]
           theta = 40.46 or 139.54 degrees included angle.
this can all be compressed and re-written as;
  cos(theta) = (sin(a)*sin(b)*sin(c))+(cos(a)*cos(b))]  
where a & b are roof angles and c is the bay window angle,
(c = the rotation out of the rectangular bay shape).