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Posted

Hello, I purchased some beginner practice exercises, and on the attached exercise I am having trouble drawing the top LH side, positioning the R5 and R3 circles, and then the R10 arc.

 

Would anyone advise if I have enough information to do this?

 

kind regards,

 

PJ

CAD 2D-2.jpg

Posted (edited)

I would draw all three as full circles first starting with R5 and R3. When it comes time to draw R10 use the Tan-Tan-Radius option of the Circle command then use the Trim command to clean things up.

 

Before and after. Oops. Wrong radius shown at bottom left. I'll fix it. See post below.

r3r5r10.jpg

Edited by ReMark
Posted

There is no height location for the R5 nor any information for R10 location.

Posted
There is no height location for the R5 nor any information for R10 location.

 

I had no trouble laying it out. I started at the circle with the radius of 3 and worked clockwise.

 

Oops. Just realized I used the wrong radius at bottom left corner.

 

OK. Fixed it and added dims.

 

r3r5r10fixed.jpg

 

I used a metric template file hence the symbol instead of the symbol.

Posted

The centre co-ordinates of the R3 circle are easily determined, but the centre co-ordinates for the R5 circle are indeterminate. Once these two circles are drawn it would be easy to draw a R10 circle tangent to these, as suggested above.

 

Any suggestions on how to position the R5 circle please?

 

Also, the above drawings have different shape for the touching of the R10 arc and the R3 circle to what is shown on the Exercise drawing. It does not touch at the horizontal diameter line of the R3 circle as the replies have all shown

Posted
I had no trouble laying it out. I started at the circle with the radius of 3 and worked clockwise.

 

Oops. Just realized I used the wrong radius at bottom left corner.

 

OK. Fixed it and added dims.

 

[ATTACH]62472[/ATTACH]

 

I used a metric template file hence the symbol instead of the symbol.

 

But, you are still just guessing where the R5 and the R10 are located. The R10 is clearly not drawn on the Quadrants of the R5 and R3 as you have them

Posted

The center of R5 in the X direction is 20.4+10.8 = 31.2 - 5.0 = 26.2.

 

R5's position is fixed in the X direction but you are right, it could be higher, or it could be lower in the Y direction than what I show.

Posted

I agree that you cannot determine the co-ordinates of the centre of the R5 circle, but once the R5 and R3 circles are positioned, it is easy to do a tan-tan- radius circle to connect these

Posted
The center of R5 in the X direction is 20.4+10.8 = 31.2 - 5.0 = 26.2.

 

Thanks ReMark, but what is the vertical distance (Y direction) for the centre of the R5 circle

Posted

Thanks SLW210, this looks closest to the Exercise drawing. Can you please tell us how you did it?

Posted

Yes I can place the R5 in very many positions.

Best Guess 2.jpg

Best Guess 3.jpg

Posted

When you get back to class ask the instructor how he/she would do the same exercise.

Posted

I also think that the R5 and R10 circles are not positioned. I would say that these two circles are tangential at the Ortho points, i.e. the centres of the two circles lie on a vertical line.

 

Therefore to draw them, I would offset the R3 circle by 10, and offset the left hand vertical line A by 5. Where these intersect at X (the upper intersection point) is the centre of the R10 circle and hence the R5 circle can be drawn.

Exercise-A.jpg

Posted (edited)
I would say that these two circles are tangential at the Ortho points, i.e. the centres of the two circles lie on a vertical line.

 

Thanks eldon, this looks to be a reasonable assumption, looking at the exercise drawing, and with the lack of any more definitive information for the centre points of R5 and R10 circles.

 

I like the way you have drawn the two loci ( the red straight line and the red circle) of the R10 circle to determine the centre point.

 

And your end result looks very much like the exercise drawing.

 

Many thanks to all that helped with this problem, I have learnt lots.

Edited by PJ01
Posted

My $0.05 Like every one else need a angle of common TP or a length on left side. Its poor workmanship when these examples are missing a vital piece of information.

Posted

Sometimes, geometry is not compatible with tidy dimensions :cry:

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