I'm fairly sure I've now sorted this.
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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!
cad.jpg
I'm fairly sure I've now sorted this.
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.
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.
cad2.jpg
Also, I started off with a radius 5 circle centred on the right of the measurement 8. Should have mentioned that.
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!
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.
Layer 0 for ByBlock Block entities, everything else ByLayer. So many issues can be solved with good templates.
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.
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.
8.jpg
I've just been down graded from 2012LT to full Autocad 2017. I WANT LT BACK
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