Btw I hope that's clear enough which lines I'm referring to....I've highlted them and their end points are showing.
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I've actually got one more question if that's ok yall!
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As per the image above, if you wanted to fillet an arc between the cyan and magenta lines, the cyan being an arc and the magenta being a straight, is it possible to get an arc in there that will have its end point at the end of the magenta line? I would like that point to be the tangent point for the arc. Ive tried guessing radiuses, and I tried a circle TTR which worked I think, but I couldn't trim the circle as I was getting the "does not intersect at 2 points' message.
Thanks everyone!
Btw I hope that's clear enough which lines I'm referring to....I've highlted them and their end points are showing.
I would construct the arc. There may be an automatic way of doing things.
First of all, draw a line from the centre of the arc to its end point. Then draw the tangent to intersect with the magenta line (it is very easy to draw lines at 90° when Polar tracking is in action). Now draw the angle bisector between these two lines (blue line). Now project your first line to intersect this angle bisector line. Where they cross is the centre of the arc.
Kerb nosings are sometimes precast in 90° sections, so the actual junctions with the straights can be a bit of a bodge.
The way I construct an angle bisector is to draw a circle centred at the intersection point. Then trim this circle with the tangents. Then draw a line from the intersection point to the mid point of the arc. This is the angle bisector.
Re kerb radius it can go as low as zero for back to back if your lip is 0.6m rad with a 0.3m kerb then radius is zero. Often the ped walk is cut at 90 so kerb goes around again rad is zero with a 45 mitre.
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Given the two point options Eldon your answer is correct for matching the arc end point draw a line from end to center point then the circle tangent length then a 90 line from the line point gives the new center.
To match a known point is a bit harder and may require an iteration to keep filleting till end point matches the point co-ords. There is a mathematical answer and its inside surveyors reverse curve formulas, the one required is for non parallel lines. Its something that is on my wish list to do as its a pain in Autocad to do reverse curves very quickly. Not at work else would post diagram.
For parallel then variables are R1 R2 offset and distance between ends or combinations of known solving the unknown. These are fairly simple mathematical formulas.
A man who never made mistakes never made anything
I see that my solution only uses the end of the arc as a fixed point, but not the end of the straight. So the end of the straight is wherever the arc ends.
My feeling is that it is not possible geometrically to join your two fixed points with an arc being tangential at each fixed point. You need a transition curve of varying radius. Or you could draw a polyline and give it tangential directions at each end.
Your correct Eldon it comes down to picking a radius, or which point to match to. The attached is the formula for reverse curves.
A man who never made mistakes never made anything
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Thanks guys for sharing your thoughts, appreciate it. I never gave arcs and curves much thought, ran away from them, but understanding them is so important for horizontal geometry!
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