# Creating a curve with radius 5 circles

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If drawing two radius 5 curves, so that the horizontal is 8 units as shown, where would you place the centres of the 5 radius circles?

I think possibly also a circle may have to be offset to achieve this, I'm not sure.

I've tried various scenarios of radius 5 circles, but can't get it to quite match.

Thanks for any help!

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I'm fairly sure I've now sorted this.

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To get all circles to be tangential, there is only one solution to your problem, given the required circle radius, and it determines what is the vertical dimension between the lines.

As you have solved it, perhaps you could share your method for the benefit of others.

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The way I went about it was as follows.

I drew a radius 10 circle centred on the right side of the measurement 8.

Then I drew a radius 5 circle centred on the left side of the measurement 8.

Then where this last circle and the radius 10 circle intersect I drew a new circle of radius 5 (highlighted in image below.

Then I trimmed away excess.

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Also, I started off with a radius 5 circle centred on the right of the measurement 8. Should have mentioned that.

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Thank you for sharing your solution.

It is pretty much how I drew it up, but, as ever, there are many ways that lead to the same solution, although some use fewer key strokes than others!

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Too much information. For the arcs to be tangent to lines and each other you cannot hold both the 5' radius and the horizontal distance of 8'. The image above holds only the 5' radius.

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I would have thought just enough information. For a horizontal distance of 8, there are arcs of many radii to join, but each arc has its vertical distance.

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The question implies that the distance of 8 units is the distance between the centres of the two arcs measured horizontally, it also implies that the where the arcs meet they are tangent. With only that information there are 2 answers.Both of which use a vertical difference of 6 units between the centres. Though obviously, only one meets the visual image given.

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The question implies that the distance of 8 units is the distance between the centres of the two arcs measured horizontally, it also implies that the where the arcs meet they are tangent..........

There are in fact a couple more answers (what is a negative sign between friends), but scenario 1 seems to meets the OP's requirement.

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• 3 weeks later...

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?

Edited by L-P
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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

:

Specify end point of arc: @4,2

Command: a

ARC Specify start point of arc or

:

Specify end point of arc: @4,2

Command: l

LINE Specify first point:

Length of line:

Specify next point or [undo]:

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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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