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I was having a look again at the replies by eldon and steven-g above.

From my attempt, shown in the 2nd image I posted above, the vertical height of the curve I drew would be 5 since the circles were radius 5.

Are you both saying that the vertical height should be 4?

How do I position the radius 5 circles to achieve the height of 4? Or do I use one of the tan functions within AutoCad to achieve this?

2. The initial requirement was for a reverse curve using curves of radius 5, with a horizontal separation of the tangent points being 8. There is only one solution for this and the vertical separation of the tangent points just happens to be 4.

To construct this, I first of all drew the line A. Offset this upwards by 5, and draw line B to join the end at C. Draw a circle radius 5 centred on C, and offset this outwards by 5 to give circle D. Offset the line B by 8 to the right, and extend it downwards to meet circle D at E. This is the centre of your second 5 radius circle. Measure up 5 from E to give the second horizontal line. This happens to be 4 higher than the first horizontal line. Draw in the circles and horizontal line, trim and there you are.

However, I noticed that the difference in coordinates were nice small integers, and using the fact that AutoCAD will draw an arc tangential to a line or arc drawn immediately before, I drew the lines and reverse curve directly with this as the command line:-

LINE Specify first point:
Specify next point or [Undo]:
Specify next point or [Undo]:

Command: a
ARC Specify start point of arc or [Center]:

Specify end point of arc: @4,2

Command: a
ARC Specify start point of arc or [Center]:

Specify end point of arc: @4,2

Command: l
LINE Specify first point:
Length of line:
Specify next point or [Undo]:

3. Thanks very much for that. I tried that for myself using your first method.

Interesting that it does give a different solution to my original attempt.

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r1=r2 = (H^2 + dist^2)/4H

0=H^2-r*H+dist^2 solve for H

If r=5 H=4

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