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Replies
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I am currently rebuilding the stage for a local theater company. The basic stage is 60'x20'. The artistic director wants an arc, whose chord is 60', and at midpoint 7' from the stage, framedand finished with a 1x10. The radius of this circle would put me somewhere in the creek behind the stage and not accessible. Any ideas on how to mark the joists that will run wild?
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Jim - there has to be a way to determine how much length each joist looses as you move from the middle one toward each side. A trigonometry teacher should be able to tell you. As long as you are spacing them evenly, it should be a fairly simple trig problem. Have him/her post it as the "problem of the day". - yb
*OK, i think it came to me while shaving. But I am assuming when you say the "cord" is 60' you mean the radius. Or you could mean the diameter, I have never seen that term used in relation to circles. Either way, you'll be able to adapt this. Now if I can "say" it...ok, divide everything by 10 to get it to a workable scale. So instead of a 60' radius (or diameter) draw a circle on a floor with a 6' radius (or diameter) Right about now I would also start thinking in inches, not feet and inches.Now- find the center of that circle, and draw a straight line through that center, bisecting the circle. Draw another line perpendicular to the first bisecting line, again through the center point.You will now have a circle with 2 lines bisecting it through its center, perpendicular to each other. Let's call them lines "a" and "b" Think of line "a" as the center joist. Draw a series of lines parallel to this line spaced 1.333" (16" divided by 10). These parallel lines have to extend through the circle.Now measure from the line "b" to the point where each of the parallel lines intersects the circle. Multiply this measurement by ten. That should be the length of the coresponding joist. (from a line you will have to draw on the stage)I am sure this will work. If I have totally confused you with my words, e mail me and I will fax you a drawing that will more clearly illustrate what I mean. Good luck, and Great question! - yb
*I suggest you do a scale drawing. If you have CADD it will be easy. Draw the cicle through your 3 points. Draw the joists and scale the lenght of each. If you don't have CADD your radius is 67' 9.5".
*Jim: I think you mean chord when you say chord. And I agree with Mike that the radius of this circle is 67.7857' or 67' 9.4286" or about 67' 9 and 7/16". Put a joist on the center line (that will project 7 feet past the cord). You know the spacing of each joist to each side (1.333', 2.666', 4.000', etc.) That is Y in a right triangle. R is the radius, 67.7857'. You get X which is the square root of (R^2-Y^2). Actually your overhang is X-60.7857'. Where that 60.7857' is the distance of the circle center from the chord ( R minus 7 feet). Punchline: For any joist Y feet from the centerline, the overhang is SQRT(67.7857^2-Y^2)-60.7857. Ta da! Plug that into a spreadsheet and have it spit out overhangs for 0, 1.333, 2.666, 4.000, up to 29.333 feet. Or plug in your particulr joist spacing. Attached is an excel spreadsheet for 16-inch spacing. Note that at the center, the overhang is 7 feet and at 30' to the side, the overhang is 0 feet. No CADD needed, just high school geometry and a calculator. -David
*Jim,
Joseph FuscoView Image
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Sorry Joe,
I think your method will produce 2 straight lines, not an arc. I've seen enough of your work to know you would've caught this right away. John
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David - what is a "chord"? thanks - yb
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Sorry Joe,
I have to agree with David Thomas. The formula you are suggesting will create a triangle.
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Bob: A chord is a line segment connecting any two points on the perimeter of a circle. The diameter is a one kind of chord, the longest one possible, between two points 180 degrees apart. Take a circle and cut off any hunk with a straight line. That cut is a chord. -David
*David's forumlas look correct to me (just got through drawing the picture).BTW, the angle of the cut at the end of each joist is ARCCOS(Y/R), e.g., 90 degrees for the joist at the midpoint, and 64.36 degrees for the joists near the endpoints (29'4" OC from the midpoint)."chord" is the straightline distance between the two endpoints of an arc.
*Thanks - after I asked you, I went back to work, and I'm thinkin', and workin', and thinkin' (and smoke is startin' to apear) about that word "chord". Of course the chords of a truss come to mind, and I then thought maybe that was what it was. Still haven't the foggiest how you got from the info given to the radius though. Maybe it will come to me. Thanks again - yb
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Joseph FuscoView Image
*John,
Joseph FuscoView Image
*gregn,
Joseph FuscoView Image
*A constant difference between joist produces two triangles, just like the studs in a gable. To make an arc, the difference between joists must increase as you get further away from the midline.
*7' minus 5/32" is the correct overhang for the joists 16" each side of center. But subsequent joists are shorter by an ever increasing amount. When I drew the larger angles, I realized that the algebra/geometry (X^2 + Y^2 = R^2) applied. You could have a construction calculator spit out the run because you know the rise (16", 32", etc.) and the diagonal (R=67.7857'). Probably what you did, but you then have to input a new rise (and same diagonal) to solve for each new run.What's bugging me now is that I tried to do this on my slide rule (K&E Model 4080-3 Log Log Duplex Trig) and I can't use the above X^2 + Y^2 = R^2 method because the significant figures aren't enough with such a long radius. That's annoying. I can get the sine and the arcsine, 1° 7.8" (should actually be 1° 7.62" but slide rules are always approximate) Then I flail on taking the cosine of that angle. Operator error. They got to the moon using these things and built Norte Dame with less but it's a dying art. Of course in the middle ages, you'd just have the serfs build a platform on the other side of the river if that's where the center is, cut down all the trees in between and then strike an arc using a long string. Speaking of medieval techniques, here's one from lofting boat planks: Take a 62.155 foot (or longer) board (really tricky without splicing several together) and bend it so it lands on the two corners of the stage and at 7 feet out in the center. 3 points on the perimeter define a circle and a uniform material will take a constant curvature. That 62' 1-7/8" figure is how much facia board you will need (52.54° angle / 360° x 67.7857 radius x 2 x pi). This wouldn't be as exact as doing the algebra (because the board isn't perfectly uniform) but it would be close, look okay to the eye, and doesn't require batteries. -David
*BarryO,
Joseph FuscoView Image
*A bit off topic but, since designing theatres is what I do for remuneration, I'd like to point out I seldom design or recommend a curved edge to a permanent stage since eventually after they open - sometimnes within days - they want to build a stage extension - either removable or semi-permanent. And its one heck of a problem and waste of time to build reverse curve platforms that fit tight and snug. Recommend a straight edge in teh center - 7' from the chord line - maybe 24' long (like 3-8' pcs of ply for the stage extension?) and taper back on a straight line from there. You'd be doing t
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Joseph FuscoView Image
*David,
Joseph FuscoView Image
*Bob: To get the radius, I considered one half of the stage: The 7 foot overhang is on the circle as is the far edge of the stage, 30 feet to one side. The center of the circle is straight behind the center some distance, X. There's a right triangle with the following corners: 1)center of circle, 2)edge of stage, and 3)intersection of centerline and chord. That triangle has a hypotenuese of X, one side (rise) of 30 feet, and a the other side (run) of X-7. So X^2 = 30^2 + (X-7)^2. Expanding that gives: X^2 = 900 + X^2 - 14X + 49. Cancelling the X^2's and adding 14X to each side gives: 14X = 949. X=949/14 = 67.7857. Or 67.76 using my slide rule. :)
*Joe: I admit a differential equation would work, but even O.D.E's make my brain hurt. Geometry and algebra are fun. I use them when I can. Newton's contributions were 1) a huge advance in mathematics and 2)usable by me 20 years ago. Whereas LaPlace, LaGrange and the other dead French guys made for a gruelling semester for me. (The 7-year absence while laying pipe and hanging rock didn't help). -David
*David,
Joseph FuscoView Image
*Joe: I'm pretty sure the formula works if all numbers are in the same units. My Excel table agrees exactly with your calculation for the first joist to the side, 5.036/32 of an inch shorter than the 7-foot overhang. So, unless Jim's using a light saber accurate to 1/1000 of an inch, call it 5/32" even. -David
*I get it. Nice work. - yb
*David,
Joseph Fusco View Image
*David, FIW, the trick to handling numbers of any size on a slide rule is to convert them to scientific notation, eg. 673 becomes 6.73 x 10^2. you then handle the mantissas on the stick and the exponents by hand. A major pain in the butt, but that was life before scientific calculators. Are you a ludite, or just a fan of antique technology?
*Joe - I know you don't care, but I want to tell you, you showed me something right there. Nice. - yb
*Daivd's formula is correct. If you plug in the spacing for the first joist on either side of the 7' center joist (i.e. 16" or 1.33'), you get the length difference Joe cited. You then have to reapply the formula for the next joist, which is spaced 32" or 2.66' from the center joist, etc, etc. The joists are just segments of chords perpendicular to the original 60' chord.
*yb,
Joseph Fusco View Image
*JT: Could there be anyone at the address [email protected]? Seems like an oxymoron. Like an old brass spirit level that is still true, I appreciare the fit and finish of high-end old tools. That it still works great after 40 years and can never have a Y2K crash makes it more elegant to my eye. It's not the exponent that flummoxed me. (I think the great value of slide rules in education is that they force you to maintain a sense of scale or you'll be off by (at least) one order of magnitude). Rather that the answer (5/32" for the first joist) is the difference between SQRT(4594.90) and SQRT(4593.12). Or 67.7857 versus 67.7726. I've only got a 12-inch slide rule so I'm solid on the first 3 sig figs but the fourth is sketchy. They did make drum slide rules with the number line segmented into a score of scales. Gave accuracy to 4 or 5 places. Beyond that, it's either paper and pencil or electrons. -David
*Joe: Accepted. Thanks. While others didn't understand your wording, I saw that your calc was smack on. If you and I could make pencil marks on the stage (like on a real job site), we'd make the identical cut. But the terms and langauge had us butting heads again. Maybe that's why we earn our livings building things and not as poets. -David
*David,
Joseph Fusco View Image
*First of all, I need a head count of all of you who went to college. Then I need to know which ones majored in some sort of advanced mathematics.Guess I would have been across the river, knocking down trees to run my line.Joe, I'll bet a weeks pay that you are actually less fun to work for than I am.Pete Draganic
*Pete,
Joseph Fusco View Image
*No, Bill, this is a time-honored technique in many fields. If you don't feel like solving the client's question, just tell them that they're asking the wrong question!Think Pete and Bill have the KISS principle down right. Draw the arc somewhere that you do have the space you need. Mark it on a wide strip(s) of paper, like butcher paper, then bring this template inside to mark off the arc on the joists and plywood. Or save a step and carefully arrange all your plywood outside and scribe directly on it.
*The problem is easy to solve and not hard to do and since I get paid for the design either way, its hardly a time honored technique from my perspective. In fact, I'm really suggesting that the KISS principal should be applied starting in design - not just in the building.
*Actually "don't feel like" were the wrong words -- more like "think the whole idea isn't really what the client needs" or some such. I think most people who take pride in their work don't view themselves as robots and don't want to design or build something that they think isn't the best solution. Of course, if the clients insist and will pay for it, it is always up to them, short of dangerous conditions.Please tell me that you insist on adequate legroom!
*Well, I'm one who recommends a slightly wider seats than average and slightly greater back-to-back than average - but there is always the management desire to squeeze in as many seats as possible - so somtimes I do lay it out the way the owner dictates (as long as it meets code minimums).
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Jim,
I believe Dave Thomas's answer is correct (Sorry Joe). Dave's formula was hard to follow so hears an easy one. I got this formula for solving radius's from page 32 of THE WOODWORKER'S PROBLEM SOLVER, By Tony O'Malley. Here's the formula:
Radius =(Length divided by 2)squared + (Height squared).......and this total is to be divided by the the sum of (2 x Height).
An easier way to show this formula is:
R=(L/2)2 + (H)2
_________________
2H
Your chord of 60' is the "length" and your 7' extension at midpoint is the "height". By using this formula, the answer computes to a radius of 67.785.....which would be 67ft and approx. 9-3/8 inches. If you don't have the room to lay this stage out inside, then your only other options are to lay it out full size somewhere outside or butt some plywood together to make a 6ft square, and lay it out from there. AS was mentioned earlier, convert your plywood into a 1inch = 1 ft. scale and mark your radius appropiately. Draw a horizontal line 60" long (represents the 60' chord) and a vertical line right at midpoint. Extend this vertical line 7 inches "North" of the horizontal line (this represents the 7' stage projection). At this point, measure back along this vertical line for a total distance of 67.785 inches and mark this spot. Now cut a stick to the same 67.785 inch length and attach it at that last spot you marked. From this vertical line, space off evenly other vertical lines which are to represent your joists. If 16"OC, space lines 1.33 inches apart from each other. If 2'OC, space them 2 inches, etc. Let all these lines run "wild" past the 7 inch midpoint line. Now use your tramel like a compass and mark each joist's position. Measure the distance from the base of the horizontal line up to where the tramel marked each joist. This answer in inches will give you the answer in feet.The only trouble with laying out something on this small of a scale, is that when the tramel passes over each joist, the joist would actually have a "long point" and a "short point" since the joists are to be cut in a "rounded" fashion and not simply cut off in a "straight butt". Since this scaled drawing would only use a small line to represent each joist, the "long point" and "short point" will be hard to determine. Oh well. Good luck with it.
[email protected]
*Jim, Sorry about that mathmatical equation, it should have looked like this: R=(L/2)2 + H2 --------------- 2HDavo
*AW SHOOT!I want that last part of the equation, "2H" to be positioned underneath the rest of the formula, but for some reason (me I guess) the computer initially shows it the right way, but when I click the "Post Message" button, it comes out wrong. Well here's another way sportsfans:(R=(L/2)2 + (H)2) / 2H Hope you get my drift. Davo
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davo,
If you need to go down a space and the computer wants to keep placing it where it wants to, then type this at the end of the line where you would like to go to the next line....
BR and it needs to be put inside of these <> i cant put them together here because it will read it as a command and not show it but just give me a space for it instead. What I mean by a space is what you get each time you hit the enter key when you type.
This message board understands html programing so sometimes it does move things around a bit by itself if you don't instruvt it to do otherwise bvia html commands. You can change text color and size like this. and as you should know you can add pictures and sounds.
Now then, I hope that is unclear enough for you.
Pete Draganic
*A brute force solution would be to create the formula in a paint or draw program, then save it as a jpeg. Post the resulting image, no HTML coding needed.Ergo (ran out of ink towards the right...):View Image
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Joseph Fusco View Image
*I really think we all could have built this stage ten times by now!!!!! Time is money...honey!]Near the stream and scaling then freehanding it till it looks like a beauty!~Near the stream~J
*Jim,For your architects arc to start at zero on the sides of a 60' wide stage and describe a 60' radius, the center joist will have to overhang the stage (Including fascia) 8.0384 feet. The old pythagoream theorem x sguared plus y squared equals hypotandsnooze square. The hypotandsnooze will always be 60' (the way I read the directions). X will be 16",32",48" to 30 feet each side of your center line.60 squared = 3600 minus 30 squared = 900, Answer =2700 Square root of 2700 = 51.961 Subtract that from 60= 8.03 feet.Doing the pythagorem over and over using the framers litany will get you joist lenghts.Tell the architect to check his math.Clampman
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Jim, the chord of a section of circular arc is the straight line from one point on the circumference of the circle to another. In other words, what the fellow in the original question wanted was a portion of a circle where the end of curve to end of curve distance (in a straight line) was 60'.
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