Newby arch question
Newby arch question
(OP)
I'm building an arch foot bridge over a small creek using large concrete blocks, but I've decided to reduce the elevation by using an ellipse. My question is regarding the calculation of the segment angles. In a normal arch the span is divided into sections and each block is shaped into a isosceles trapezoid. The side angles are derived by taking the 180deg arc and dividing it into the desired number of segment sections and bisecting these points through the focal point of the radius. The isosceles trapezoid causes a wedging action with its neighbor giving compressive strength...but you probably aready knew that :)
How do I do this with an ellipse? Where is the best height of the center point for the angles to start from for wedging to occure?
How do I do this with an ellipse? Where is the best height of the center point for the angles to start from for wedging to occure?






RE: Newby arch question
BA
RE: Newby arch question
Look at the "Five Center Method" at this site:
http://www.uwgb.edu/dutchs/MATHALGO/Ellipses.HTM
It's a leftover from my draughting years.
Show us your result.
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Newby arch question
Sorry about that.
As BA says, you can go back to a circular arc. With abutments that can resist horizontal force, you don't need the full half circle.
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Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Newby arch question
A flat arch is going to have horizontal forces at the abutments no matter whether it is elliptical or circular.
The circular segment is probably better for stability because the transition from a tangential force to a vertical force takes place in the abutment, rather than in the arch.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/