Hope I can illustrate this verbally:
You draw a line. Say its 15″ in length. Now you want to draw a circle such that if the line were to go from edge to edge of the circle, the point where the ends of the line intersected the edge of the circle, the circle would be 15deg off the straight line. On both sides. So how to you figure how big that circle is?
I thought I could do it in 10 seconds with SU. Naaah.
Real trucks dont have sparkplugs
A new code-compliant, spill-safe outlet from Legrand offers a sleek solution for a kitchen island plug.
"I have learned so much thanks to the searchable articles on the FHB website. I can confidently say that I expect to be a life-long subscriber." - M.K.
Dig into cutting-edge approaches and decades of proven solutions with total access to our experts and tradespeople.
Start Free Trial NowGet instant access to the latest developments in green building, research, and reports from the field.
Start Free Trial Now
Replies
Where I get lost is that a circle does not have an edge. It is continuous.
And a circle does not intersect a line at any angle. It takes two straight lines to vector an intersect that can be measured in degrees.
Got a sktch to show what you mean?
Welcome to the
Taunton University of Knowledge FHB Campus at Breaktime.
where ...
Excellence is its own reward!
I knew I'd muck it up trying to put it in words
The 15 deg is probably not an absolute . . . circles have curves. But like a tangent line is tangent at precisely only one place . . . think in those terms
Maybe this helps?
Maybe if I tried to make it a construction problem. You have a board in front of you. It has to have a curved part made to fit it. You can't cut or change it, but at present it has a 15 deg cut on the each end (really capable miter saw, there). You know how to use a router and a circle jig, you just dont know where to set the pin so that they meet up nice. Real trucks dont have sparkplugs
Maybe I just figured it. I just told SU to scale it so that line was 15", and I got r= 19 5/8".
But I dont have SU in my pocket all day. I'm still interested in the math part. Real trucks dont have sparkplugs
I agree with Paul & Joe. The radius is 28.977".I tried to solve it without using the method that I employ to determine the radius of an arch when only the width & height is known.I could not so I stuck with old faithful. Radius = (H² + .25{2W²})/2 * HH= 3.8823"
W= 28.977"
R= 28.977"The CAD drawing attached gives a visual reference.
Jon Blakemore RappahannockINC.com Fredericksburg, VA
Does this help?
I'm thinking that the 15" dimension is where you have the green line segment not 15" as the diameter.
Assuming Oldguy is right that the green line is the 15" line, then the radius of the circle is (7.5/cos 15) = 7.76"
edit to correct typo
"Listen, strange women lying in ponds distributing swords is no basis for a system of government."Jon
Edited 10/15/2007 8:58 pm ET by Hector45
Yes, it helps. Consider the old right triangle situation. The length of the third side of a right triangle is the square root of the sum of the squares of the other two sides. In this case, you have what? 7.5"? for each of these sides? So, the third side would be 10.6066", but that's for a 90 degree angle. We want a 15 degree angle. 90 / 15 = 6. So, divide 10.6066 by 6 and you get 1.767767. Draw a triangle 7.5 x 7.5 x 1.767767, and it will be a 15 degree angle.Another approach would be to use trigonometry (sine & cosine), but it's been 30 years since I touched that stuff.George Patterson
RW, judging from your circle.jpg drawing I think what you might be after is being discussed in a recent Bowed Bay Window Roof Framing thread at the JLC Forum.
If this is in essence the intent of your question ... I've found it's one of the most challenging posed in the forums of late and there doesn't seem to be a pat answer to this one.
Here's a link to a Bow Bay Window Polygon Solver … there are links to similar calculators which work on different principles and diagrams. The "different principles" in question being what I call "dummy code"; I’m using brute force cave man style programming to crunch the numbers.
I haven’t posted the diagrams and algebra as yet … there are answers that may be obtained using math, such as Newton’s Method and trigonometry. The trouble is that the both the programs and formulas for field calculations (where, as you so well put it, we can't have SketchUp in our pockets) become more complicated. So the best I've come up with so far are the brute force methods linked to in the JLC Forum.Joe Bartok
O.K., tune out my last post, we don't need all that. I missed the 15" chord length but I guess I'm still not understanding the question.
For a 15 chord, half the chord length is 7.5.
7.5" ÷ sin 15° = 28.977775" ... different than the SketchUp result.
Anyone shed some light on this???
Edited ... wait for clarification ...
Joe Bartok
Edited 10/16/2007 10:46 am ET by JoeBartok
Edited 10/16/2007 10:46 am ET by JoeBartok
Edited 10/16/2007 10:52 am ET by JoeBartok
Edited 10/16/2007 10:53 am ET by JoeBartok
my calculations agree with yours
the radius is 28.997"
I wasn't understanding if it's one 15" line at 15° coming off the point, or is it two?
RW wrote "the circle would be 15deg off the straight line. On both sides."
That sounds like 15° on each side so we have two chords, each 15" long. That's what I'm basing the calculation on.
Another way to do this would be to think of this as an arch:
Width = 2 ×15" × cos 15° = 28.977775"
Height = 15" × sin 15° = 3.882286"
Solving the radius ... 28.977775"Joe Bartok
I keep reading this over.
If I make some assumptions, The 15" line would be the chord of a circle, and the 15° vector would be created by a line tangent to the circle. Is that right?
It is posssible that the givens creat an impossible whenn you defien the circle that way
Welcome to the
Taunton University of Knowledge FHB Campus at Breaktime.
where ...
Excellence is its own reward!
You have enough info to find the radius of the circle, cant you make your circle from this?
Or, I'm missing something very simple here.
Doug
I'm not sure I'm following your description.
Here is my favorite online calc for circle problems --- does this help?
“The Hand of providence has been so conspicuous in all this, that he must be worse than an infidel that lacks faith, and more than wicked, that has not gratitude enough to acknowledge his obligations.” —George Washington
If I am understanding your description:
7.5 divided by the cosine of 15-degrees is the radius. The radius multiplied by the sine of 15-degrees gives you the offset between the line and the center of the circle.
I believe you're describing a chord which is 15 deg. from the tangent at each end. That's a 30 degree arc, so if you draw to radii from the ends of the chord, they form a 30 degree angle, so the radius/chord angle is 75 deg on each corner, etc.
r * cos 75 = 7.5
r = 28.978