# Drawing a part radius with limited information

Is there any way to draw the Radius "R4" with only the information supplied
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Commented:
I believe you would need more information, such as the arc length or intersect point(s).

This is an unusual question for the Microsoft Access topic area.  Did you intend to place the question under a different topic, or are you using MS Access/VBA to do this?
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Commented:
Mbizup is correct.
Rotate your figure clockwise until the line between the center of a circle and upper right point of your figure become vertical.
Imagine a set of circles with different sizes falling down to the angle between sides of your figure. Any circle will have 2 contact points, so there is no solution without additional data.
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Commented:
Are you trying to calculate R4, or are you saying it's a given, and you just want to know how to plot the points of it's arc for the "rounded corner"?  If you are trying to calculate R4 then the answer is no.  There are many circles of different radius that will round the corner at different "sharpnesses", and meet the criteria of being tangent to two lines that are the intersecting sides of your figure.  For example:

»bp
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Commented:
Even if "R4" is a known/fixed dimension, you would need additional information as to where the arc is in relation to the trapezoid.
(Cut an edge off a square post-it note, similar to your picture, and place a mug or other round-based object on it with the center at various distances from the sides to see the dilemma.)
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Commented:
Hmmm, I think if R4 is known (the radius of the arc / circle), then the center can be located.  Drawing parallel lines to the existing intersecting sides at distance R4 from those lines, then where those two new lines intersect is the center of the circle / arc.

»bp
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Commented:
<<parallel lines to the existing intersecting sides at distance R4 from those lines>>

We know the radius of the circle (R4)... but the original picture doesn't explicitly show the relationship between the circle and square/trapezoid (whether the circle is entirely contained, what portion of the arc intersects, etc).
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Commented:
Fair enough, I am going from the assumption that we want the two lines that the contained circle / arc intersects are tangent lines to the circle.

»bp
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Commented:
This is not indicated, but most of us assumed that the circle touches the lines at one point without crossing them. In this case, the center of the circle is on the bisector of the angle.
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MDAuthor Commented:
Some excellent ideas. Love the parallel idea.

This is part of a drawing requirement and need to be done in access. As any of the values can vary so I need a formula. The numbers shown would always be known.

If we could somehow calculate the length of the line indicated we would know the center of the radius.

Are arcs drawn point to point or by angle?
How do you get the bisector of the angle without knowing the angle? The angle could be calculated?
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Commented:
Is the radius of the arc you want to use a given, or to be solved for?

»bp
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MDAuthor Commented:
oops! No we would not know the center of the radius. I don't thinks so anyway.

So what is the minimum extra info that we need.?

I guess the angle of the slanted line in relation to the circle or vertical line might help?
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MDAuthor Commented:
The radius is known and all other dimensions shown are known.

There must be some trigonometry that gives the angle based on the two right angled lines?
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Commented:
I'm not adept with drawings... but from the information you have, you can derive the angle that encloses the arc.   In radians, it would be:

Pi - ARCTAN(22 / 5)

>> So what is the minimum extra info that we need.?

If you make a complete circle, is your arc guaranteed to intersect each line in exactly/only ONE place (ie: are the lines tangential to the arc)?

If so, then I think you can use trig formulas to derive the distances from the corner to the intersect points using the previously derived angle and the radius.  (Not exactly sure how, but I can dig into it if you don't get other responses)

If not, then you'd need some other piece of information to define the circle's position.
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RetiredCommented:
assuming that the 2 points that touch the circle are tangent to each side, then this is the result:the tangent angle that subtends the arc is 22/5 = 4.4

in the tan tables, tan 77.2° = 4.401...
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Commented:
You can see that small triangle is similar to the triangle with sides 5 and 22 mm (all angles are the same, so these triangles are similar).
Now we can calculate ratio:
X/SQR(5^2 + 22^2)=R4/22
X = R4/22*SQR(5^2 + 22^2)
X = R4*1,025501288425316
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MDAuthor Commented:
Thank you all great work!

can anyone derive the center of the arc I assume I could then draw the arc using mbizup Pi - ARCTAN(22 / 5) . That's if arc are drawn that way. I have only drawn complete circles before but know that there are extra values that can be added after this:

Me.Circle (1000, 2000), (500), 255, ?,?
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MDAuthor Commented:
I just read all posts again. I think I can get the arc center. I just need to know how to draw the arc
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RetiredCommented:
1. draw perpendicular lines from the points where the 2 sides touch the arc
2. at the point where these perpendicular lines intersect, this is the center of your circle which will give you an arc that touches both sides
3. using a compass, draw the circle with a radius of either of the red radii (both are the same length)
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MDAuthor Commented:
So sorry all I meant draw circle in code as example:

Me.Circle (x, y), (Rad), Colour, ?,?

As you can see there are two other parameters that can be set. I have checked help in access 2002 but it certainly is not clear
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RetiredCommented:
@Derek Brown sorry, not so good with code
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Commented:
Due to:
https://msdn.microsoft.com/en-us/library/office/aa221156(v=office.11).aspx
other parameters are:
start angle = 0 (never tried, but I hope angles are started from 12), in article above –.00000001 is recommended
end angle = 3.14159265359 - ATN(22 / 5)
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MDAuthor Commented:
I think if arcs are drawn in Radians it is 3.00 o'clock for zero start and angles seem to be drawn anti clockwise. But I am still checking.
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Commented:
If you are correct, you should use:
PI = 3.14159265359
start  angle = PI / 2 - Atn(22 / 5)
end angle = PI / 2
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MDAuthor Commented:
It looks that way but I may be drawing arc using a minus value.

Nearly have it all done thanks to your help. Will post final solution when I have solved another question that I have just posted. It effects the physical drawing that access does so cannot confirm if formula is correct. To do with scaling. It draws circles as circles regardless of scaling but does scale lines and boxes.
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MDAuthor Commented:
Hi Alsa315

Should this be:
PI = 3.14159265359
start  angle = PI / 2 - Atn(5 / 22)
end angle = PI / 2
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MDAuthor Commented:
Thanks to all, great help. The original solution that we finally got to, used trigonometry but that caused errors when the "R" (Radius) was Zero causing error "cannot divide by Zero" . So the following was created using just Pythagoras and similar triangle rules:
Dim W As Variant
Dim H As Variant
Dim R As Variant
Dim A As Variant
Dim PI As Variant
Dim C As Variant
Dim Text As Variant
Dim LL As Variant
W = 20
H = 22
R = 4
A = 5  'Top right corner to intersection of where slanted line would  meet to horizontal line
PI = 3.141592654
Me.Circle (W - A - (R * H) / Sqr(A * A + H * H) + A * R / H * (1 - A / (Sqr(A * A + H * H))), R), R, 0, Atn(A / H), PI / 2
Me.Line (W, H)-(W - A + A / H * (R - (A * R) / (Sqr(A * A + H * H))), R - A * R / (Sqr(A * A + H * H))), 0
Me.Line (W, H)-(0, H)
Me.Line (0, H)-(0, 0)
Me.Line (0, 0)-(W - A - (R * H) / Sqr(A * A + H * H) + A * R / H * (1 - A / (Sqr(A * A + H * H))), 0)
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