Balanced/Unbalanced Snow Load - Dome - ASCE 7-16 Example (2)?

[Howlyn2]

ASCE 7-16's commentary does a fine job giving snow load examples in the commentary. However, for example #: 2 the building with a 80' span and 15' rise circular arc has given me pause. Truthfully, the snow loads make sense but how does ASCE 7-16 justify the distance from 30 deg. slope (30'), and slope at the eaves = 41 deg.? They state "by geometry" and I'm having a tough time matching. Although this isn't the core of the problem, I want to understand. I've tried circular arc equations, etc. and cannot find these values. What equations is ASCE 7-16 using to find these distances and angles? Thank you.

 

[ProgrammingPE]

You just need to determine the radius of the roof and the angle at the eaves to get the slope at any point. The eave x-coordinate is R*sinθ, and it is 40 feet from the crown. Its y-coordinate is R*cosθ, and it is 15 feet below the crown, so you have two equations:

R*sinθ = 40'​

R - R*cosθ = 15'​

This results in R = 60.8' and θ = 41.1°

The angle, θ, is also the roof slope (it is 90° from the radius line), so that gets you your slope at the eaves. The x-coordinate where the slope is 30° will then be at R*sin(30°) = 30 feet.

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[HTURKAK]

There are other methods to calculate the slope angle at the eaves .

Assume the center of the roof dome circle at origin . The formula of the circle of the dome;

X^2 +Y^2= 60.8^2

The derivative of the circle at any point is also the slope of the tangent line at that point.

so dy/dx= - 2x/2y= -x/y

The coordinates of the eave will be = 40 , (60.8-15=45.8)

and the slope dy/dx=40/45.8=0.873 ( slope is the TAN of the angle)

ATAN(0.873)=41.13 degr.