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#1
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Compound miter brainteaser
Not for those who don't remember their trigonometry:
My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, with a zero degree tilt (vertical cut). If the roof was infinitely steep (like building the walls of a ten-sided tower), he would have to have cut the boards with a zero degree miter (perfectly perpendicular cross-cut) with a 36 degree tilt from vertical. What would the formula be for N sides with a roof pitch of A degrees? Josh |
#2
Posted to rec.woodworking
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Compound miter brainteaser
5/12 pitch is 5 divided by 12 x 90 = degrees. 37.5
"Josh" wrote in message oups.com... Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, with a zero degree tilt (vertical cut). If the roof was infinitely steep (like building the walls of a ten-sided tower), he would have to have cut the boards with a zero degree miter (perfectly perpendicular cross-cut) with a 36 degree tilt from vertical. What would the formula be for N sides with a roof pitch of A degrees? Josh |
#3
Posted to rec.woodworking
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Compound miter brainteaser
"Doug Schultz" wrote:
5/12 pitch is 5 divided by 12 x 90 = degrees. 37.5 So 12/12 pitch, which most of us think is a 45 degree angle, is really 12/12*90 = 90 degrees? No, roof pitch refers to "rise over run". except for special cases, you are going to have to resort to tables or trig [angle = atan(rise/run)] to get the angle. Josh stated this angle correctly. I think he forgot to divide by two in his other angle. "Josh" wrote in message roups.com... Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, with a zero degree tilt (vertical cut). If the roof was infinitely steep (like building the walls of a ten-sided tower), he would have to have cut the boards with a zero degree miter (perfectly perpendicular cross-cut) with a 36 degree tilt from vertical. What would the formula be for N sides with a roof pitch of A degrees? Josh -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#4
Posted to rec.woodworking
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Compound miter brainteaser
"Josh" wrote:
Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, with a zero degree tilt (vertical cut). Josh, I think you forgot to divide by two here. each angle of an equilateral decagon is 144 degrees. To cut _one_ board to get that angle, you would cut 36 degrees from perpendicular. But presuming that you want the bevel or miter to line up (and maybe my assumption is not correct?), you would want to cut 18 degrees on each board. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#5
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Compound miter brainteaser
Yes. My bad. I meant to say 18 degrees for both of my examples.
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#6
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Compound miter brainteaser
"Josh" wrote:
Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? Answer 1: 18 degrees, after first building a jig to hold the planks at a 5:12 slope in his chop saw, RAS, etc. g Answer 2: 16.7 degree miter and 7.1 degree bevel. If that's not right, then I'll have to solve it on something bigger than a post-it, and write out my steps more carefully! If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, with a zero degree tilt (vertical cut). If the roof was infinitely steep (like building the walls of a ten-sided tower), he would have to have cut the boards with a zero degree miter (perfectly perpendicular cross-cut) with a 36 degree tilt from vertical. What would the formula be for N sides with a roof pitch of A degrees? This is the point where the prof says "The generalization is trivial, and is left as an exercise for the class." -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#7
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Compound miter brainteaser
I agree with you on the miter, not the bevel (though you're close).
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#8
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Compound miter brainteaser
Josh wrote:
I agree with you on the miter, not the bevel (though you're close). I got 8.35287747524613872127 er -- email not valid |
#9
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Compound miter brainteaser
alexy wrote:
"Josh" wrote: Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? Answer 1: 18 degrees, after first building a jig to hold the planks at a 5:12 slope in his chop saw, RAS, etc. g Answer 2: 16.7 degree miter and 7.1 degree bevel. If that's not right, then I'll have to solve it on something bigger than a post-it, and write out my steps more carefully! If you use Un*x. bc has a very limited set of functions, and then only if you invoke it with the -l flag. The initial stuff adds a tangent and arcsin function to make up for that, and defines pi so I didn't have to write it out. Save the below to a file named compound-miter.bc, and use "bc -ql [path/to/]compound-mitre.bc" to calculate the angles for a (roof) of an arbitrary number of sides and roof pitch. =8---------------------------------------- define asin (k) { return a( k / sqrt(1 - sqrt(k)) ); } define tan (p) { return s(p) / c(p); } pi=4.0 * a(1); print "how many sides?: "; #angle=360 / angle * pi / 180; angle=2 * pi / read (); # won't take a fraction needs a decimal number print "what is the slope?: "; slope=a(read()); miter=a( c(slope) * tan(angle/2.0)); bevel=asin( s(slope) * s(angle/2.0) ); scale=2; print "\nThe miter is: ", miter * 180.0 / pi, " degrees.\n"; print "The bevel is: ", bevel * 180.0 / pi, " degrees.\n"; quit; =8---------------------------------------- er -- email not valid |
#10
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Compound miter brainteaser
"Josh" writes: What would the formula be for N sides with a roof pitch of A degrees? There's a reason why I have a web page to do this: http://www.delorie.com/wood/compound...=10&angle=5/12 http://www.delorie.com/wood/compound-cuts.html Number of sides: 10 Angle of sides: 22.6 Cross Cut Angle: 16.7 Blade Angle: 6.8 |
#11
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Compound miter brainteaser
DJ Delorie wrote:
"Josh" writes: What would the formula be for N sides with a roof pitch of A degrees? There's a reason why I have a web page to do this: http://www.delorie.com/wood/compound...=10&angle=5/12 http://www.delorie.com/wood/compound-cuts.html Number of sides: 10 Angle of sides: 22.6 Cross Cut Angle: 16.7 Blade Angle: 6.8 Now three people have three different answers for the bevel. Yours is closed source, so no way to check. fwiw, I got my formula from this page: http://www.woodcentral.com/bparticle..._formula.shtml and tossed up the (interactive!) script myself. er -- email not valid |
#12
Posted to rec.woodworking
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Compound miter brainteaser
DJ Delorie wrote:
There's a reason why I have a web page to do this: http://www.delorie.com/wood/compound...=10&angle=5/12 http://www.delorie.com/wood/compound-cuts.html *Looks again* Heh. The more I look at your web pages the more silly I feel having made that statement. er -- email not valid |
#13
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Compound miter brainteaser
Enoch Root writes: Yours is closed source, so no way to check. Ok, here it is, check it. "woodlib.pl" just sets $in{} from the CGI variables. #!/usr/bin/perl # -*- perl -*- use POSIX; $pi = atan2(1,1) * 4; $original_email = ' From: "Robert Smith" Subject: Best CAD Program for Woodworkers Newsgroups: rec.woodworking Date: Fri, 31 Oct 2003 13:51:59 GMT Now to try and make the formula easier to understand we need to define two variables. The number of sides will be in variable "s". The angle of the sides, which we just calculated will be in variable "b" Just plug the correct values in this short formula and you will have your answer a=360/s x=arctan((cos b)*tan(a/2)) y=arcsin((sin b)*sin(a/2)) The "x" value will be the angle that you set your cross cut to. The "y" value will be the angle that you set your saw blade to. '; require "./woodlib.pl"; $sides = $in{'nsides'}; $angle = $in{'angle'}; $sides = 4 unless $sides 0; $angle = 0 unless $angle 0; if ($angle =~ m@([0-9\.]+)/([0-9\.]+)@) { ($rise, $run) = ($1, $2); $angle = atan2($rise, $run); } else { $angle = $angle * $pi / 180; } $a = $pi / $sides; $x = atan(cos($angle)*tan($a)); $y = asin(sin($angle)*sin($a)); print "Content-type: text/html\n\n"; print `header Compound Cut Calculations`; print "centertable"; &row("Number of sides:", $sides); &row("Angle of sides:", $angle * 180/$pi); &row("br", ""); &row("Cross Cut Angle:", $x * 180/$pi); &row("Blade Angle:", $y * 180/$pi); sub row { $v = $_[1]; if ($v =~ m@[0-9]\.@) { $v = sprintf("%.1f", $v); } print "trtd align=right nowrap$_[0] /tdtd align=righttt $v/tt/td/tr\n"; } print "/table/p"; print "pa href=\"compound-cuts.html\"Return to the Form/a/p\n"; print "/center"; print `trailer`; |
#14
Posted to rec.woodworking
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Compound miter brainteaser
Enoch Root wrote:
DJ Delorie wrote: "Josh" writes: What would the formula be for N sides with a roof pitch of A degrees? There's a reason why I have a web page to do this: http://www.delorie.com/wood/compound...=10&angle=5/12 http://www.delorie.com/wood/compound-cuts.html Number of sides: 10 Angle of sides: 22.6 Cross Cut Angle: 16.7 Blade Angle: 6.8 Now three people have three different answers for the bevel. Yours is closed source, so no way to check. fwiw, I got my formula from this page: http://www.woodcentral.com/bparticle..._formula.shtml and tossed up the (interactive!) script myself. And mine was no formulas--just visualize and figure it out, with lots of chances for errors.But sounds like we are all agreed on the miter, and just need to settle on the bevel. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#15
Posted to rec.woodworking
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Compound miter brainteaser
DJ Delorie wrote:
Enoch Root writes: Yours is closed source, so no way to check. Ok, here it is, check it. "woodlib.pl" just sets $in{} from the CGI variables. $x = atan(cos($angle)*tan($a)); $y = asin(sin($angle)*sin($a)); s/b $a/2: $x = atan(cos($angle)*tan($a/2)); $y = asin(sin($angle)*sin($a/2)); Though, I didn't verify on your page. er -- email not valid |
#16
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Compound miter brainteaser
Enoch Root writes: s/b $a/2: $x = atan(cos($angle)*tan($a/2)); $y = asin(sin($angle)*sin($a/2)); $a is already halved. It's computed as $pi / $sides, not 2*$pi / sides. |
#17
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Compound miter brainteaser
DJ Delorie wrote:
Enoch Root writes: s/b $a/2: $x = atan(cos($angle)*tan($a/2)); $y = asin(sin($angle)*sin($a/2)); $a is already halved. It's computed as $pi / $sides, not 2*$pi / sides. Right you are. It was my definition of Asin. er -- email not valid |
#18
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Compound miter brainteaser
Enoch Root wrote:
[snip] I got the definition of Asin messed up. should be: define asin (k) { return a( k / sqrt(1 - k^2) ); } er -- email not valid |
#19
Posted to rec.woodworking
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Compound miter brainteaser
alexy wrote:
Enoch Root wrote: DJ Delorie wrote: "Josh" writes: What would the formula be for N sides with a roof pitch of A degrees? There's a reason why I have a web page to do this: http://www.delorie.com/wood/compound...=10&angle=5/12 http://www.delorie.com/wood/compound-cuts.html Number of sides: 10 Angle of sides: 22.6 Cross Cut Angle: 16.7 Blade Angle: 6.8 Now three people have three different answers for the bevel. Yours is closed source, so no way to check. fwiw, I got my formula from this page: http://www.woodcentral.com/bparticle..._formula.shtml and tossed up the (interactive!) script myself. And mine was no formulas--just visualize and figure it out, with lots of chances for errors.But sounds like we are all agreed on the miter, and just need to settle on the bevel. Finally, I got 6.827. I found this pretty tricky to visualize. Glad some of you had formulas and programs, so I would know each time I got a wrong answer! Josh, are these the answers you expected? -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#20
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Compound miter brainteaser
I agree with your algorithm, DJ. Several such algorithms can be found
on the web, often for computing compound miters for crown molding, which is essentially the same problem. But many of them give slightly different answers. Why is yours right, and what's wrong with the other ones? If you don't want to see a bunch of crazy math, stop reading now. One of the other common algoriths for computing the miter (x in your notation) is x=1/2*arccos(cos^2(b)*cos(a)+sin^2(b)). This is a pretty simple equation to derive using simple vector algebra. Going back to first semester Calculus, recall that the dot product of two vectors is defined as a scalar value equal to the product of the vector magnitudes times the cosine of the angle between them. A.dot.B = |A| * |B| * cos(alpha), where alpha is the angle formed between them, and the | | notation means magnitude (i.e. length, independent of direction). A second way to calculate the dot product is to write the vectors as functions of the unit vectors i, j, and k which are simply vectors of length 1 along the x, y, and z axes. If the vectors are written as A = ax*i + ay*j + az*k and B = bx*i + by*j + bz*k then their dot product is simply A.dot.B = ax*bx + ay*by + az*bz Now if we simply find two vectors which form one wedge of the ten-sided roof, we can easily compute the angle between them by using the two definitions of dot product. It's an easy construction: If we take the peak of the roof to be the point (0,0,0), and we imagine ten rafters radiating outward, angled down with a pitch (slope) of 5/12, then it's easy to find their endpoints (which will define our vectors). If we assume for simplicity's sake that they have a length of 1 foot, then one of the rafters would stretch from (0,0,0) to (cos(atan(5/12)),0,-sin(atan(5/12))). Since the choice of 5/12 for a pitch gives us a 5-12-13 right triangle, we can simpify the second coordinate of the vector to (12/13,0,-5/13). A second rafter would start at (0,0,0) and go to (12/13*cos(36), 12/13*sin(36), -5/13). The 36 degree angle is the angle of one wedge of roof when viewed from directly above (i.e. there are ten sides so the the angle is a tenth of 360). Now that we have two vectors, we can compute their dot product both of the ways desribed above. A.dot.B = |A| * |B| * cos (angle) = 1 * 1 * cos (alpha) = cos(alpha) A.dot.B = 12/13*12/13*cos(36) + 0*12/13*sin(36) + 5/13*5/13 Equating the two different definitions we get ((12/13)^2*cos(36) + (5/13)^2)= cos (alpha) Thus alpha = arccos(0.83727) = 33.147 degrees. This is the angle between the two vectors (i.e. the two roof rafters). The miter angle is simply going to be half of this angle, or x = 16.57 degrees. This is pretty close to what DJ's formula gives us, yet it's slightly different. Why? I'll post the reason next. I don't want this one post to get too long. Josh |
#21
Posted to rec.woodworking
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Compound miter brainteaser
Even though the formula I referenced previously (which can be found as
the basis for several compound miter calculators) is easily derived, there was an essential flaw in the construction of the problem (but not in the math, itself). The crux of the problem is that even though the roof pitch is 5/12, the pitch of the rafters is not 5/12 (at least not the ten main rafters which form the wedges which comprise the roof). If we added ten more secondary rafters to the roof frame ran down the middle of each wedge, bisecting the 36 degree angle, those would have a pitch of 5/12. To put this in a way that is easier to picture, imagine that the secondary rafters are 13 feet long, thus forming a 5-12-13 triangle. The would have a rise of 5 feet, a horizontal run of 12 feet, and an overall length of 13 feet. To keep the eaves of the roof level, the main rafters would, of course, have to drop the same 5 feet total at their end points. However, they would have to be longer than 13 feet. They would in fact be 13/cos(18) feet long. Thus, their pitch would actually be less than 5/12, and that's why the other formula is wrong. Josh |
#22
Posted to rec.woodworking
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Compound miter brainteaser
somehow after atepting to try and understand
i declare myself (STUPID) "Josh" wrote in message oups.com... I agree with your algorithm, DJ. Several such algorithms can be found on the web, often for computing compound miters for crown molding, which is essentially the same problem. But many of them give slightly different answers. Why is yours right, and what's wrong with the other ones? If you don't want to see a bunch of crazy math, stop reading now. One of the other common algoriths for computing the miter (x in your notation) is x=1/2*arccos(cos^2(b)*cos(a)+sin^2(b)). This is a pretty simple equation to derive using simple vector algebra. Going back to first semester Calculus, recall that the dot product of two vectors is defined as a scalar value equal to the product of the vector magnitudes times the cosine of the angle between them. A.dot.B = |A| * |B| * cos(alpha), where alpha is the angle formed between them, and the | | notation means magnitude (i.e. length, independent of direction). A second way to calculate the dot product is to write the vectors as functions of the unit vectors i, j, and k which are simply vectors of length 1 along the x, y, and z axes. If the vectors are written as A = ax*i + ay*j + az*k and B = bx*i + by*j + bz*k then their dot product is simply A.dot.B = ax*bx + ay*by + az*bz Now if we simply find two vectors which form one wedge of the ten-sided roof, we can easily compute the angle between them by using the two definitions of dot product. It's an easy construction: If we take the peak of the roof to be the point (0,0,0), and we imagine ten rafters radiating outward, angled down with a pitch (slope) of 5/12, then it's easy to find their endpoints (which will define our vectors). If we assume for simplicity's sake that they have a length of 1 foot, then one of the rafters would stretch from (0,0,0) to (cos(atan(5/12)),0,-sin(atan(5/12))). Since the choice of 5/12 for a pitch gives us a 5-12-13 right triangle, we can simpify the second coordinate of the vector to (12/13,0,-5/13). A second rafter would start at (0,0,0) and go to (12/13*cos(36), 12/13*sin(36), -5/13). The 36 degree angle is the angle of one wedge of roof when viewed from directly above (i.e. there are ten sides so the the angle is a tenth of 360). Now that we have two vectors, we can compute their dot product both of the ways desribed above. A.dot.B = |A| * |B| * cos (angle) = 1 * 1 * cos (alpha) = cos(alpha) A.dot.B = 12/13*12/13*cos(36) + 0*12/13*sin(36) + 5/13*5/13 Equating the two different definitions we get ((12/13)^2*cos(36) + (5/13)^2)= cos (alpha) Thus alpha = arccos(0.83727) = 33.147 degrees. This is the angle between the two vectors (i.e. the two roof rafters). The miter angle is simply going to be half of this angle, or x = 16.57 degrees. This is pretty close to what DJ's formula gives us, yet it's slightly different. Why? I'll post the reason next. I don't want this one post to get too long. Josh |
#23
Posted to rec.woodworking
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Compound miter brainteaser
Josh wrote:
Even though the formula I referenced previously (which can be found as the basis for several compound miter calculators) is easily derived, there was an essential flaw in the construction of the problem (but not in the math, itself). The crux of the problem is that even though the roof pitch is 5/12, the pitch of the rafters is not 5/12 (at least not the ten main rafters which form the wedges which comprise the roof). If we added ten more secondary rafters to the roof frame ran down the middle of each wedge, bisecting the 36 degree angle, those would have a pitch of 5/12. To put this in a way that is easier to picture, imagine that the secondary rafters are 13 feet long, thus forming a 5-12-13 triangle. The would have a rise of 5 feet, a horizontal run of 12 feet, and an overall length of 13 feet. To keep the eaves of the roof level, the main rafters would, of course, have to drop the same 5 feet total at their end points. However, they would have to be longer than 13 feet. They would in fact be 13/cos(18) feet long. Thus, their pitch would actually be less than 5/12, and that's why the other formula is wrong. Hah, that's gorgeous... 5/12 is the pitch along a line perpendicular to the direction of the slope, but the rafters are not that. The other formula will work if you adjust your slope to slope = sqrt(rise^2/(rise^2+run^2)) won't it? (seems to, based on a test using my own proggie) er -- email not valid |
#25
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Compound miter brainteaser
Enoch Root writes: Hah, that's gorgeous... 5/12 is the pitch along a line perpendicular to the direction of the slope, but the rafters are not that. Except for the corner ones, my rafters are. They have to run parallel to the seams in the plywood sheathing. |
#26
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Compound miter brainteaser
"Morris Dovey" writes: You're obviously having too much fun with this. Just to open things up a bit for all the people who don't have bc (but _do_ have a C compiler): If you have neither bc nor a C compiler, my script uses the same math (and allows for rise/run too): http://www.delorie.com/wood/compound-cuts.html |
#27
Posted to rec.woodworking
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Compound miter brainteaser
DJ Delorie (in ) said:
| "Morris Dovey" writes: || You're obviously having too much fun with this. Just to open things || up a bit for all the people who don't have bc (but _do_ have a C || compiler): | | If you have neither bc nor a C compiler, my script uses the same | math (and allows for rise/run too): | | http://www.delorie.com/wood/compound-cuts.html Your script is much appreciated (but not as much your ABPW archive and nowhere near as much as your efforts to make a C compiler freely available to all) - but there are more than 15 miles between my shop and my internet access. For those with a similar situation, the C source can be found in the collection at www.iedu.com/mrd/c/ as cbevel.c - sometimes it's a Good Thing to be able to function independently from the internet. :-) -- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/ |
#28
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Compound miter brainteaser
"Josh" wrote:
Even though the formula I referenced previously (which can be found as the basis for several compound miter calculators) is easily derived, there was an essential flaw in the construction of the problem (but not in the math, itself). Well, not to be picky, but... There is no flaw in the construction of the problem. The flaw is in the solution, using the roof slope as the slope of the diagonal rafters. The crux of the problem is that even though the roof pitch is 5/12, the pitch of the rafters is not 5/12 (at least not the ten main rafters which form the wedges which comprise the roof). If we added ten more secondary rafters to the roof frame ran down the middle of each wedge, bisecting the 36 degree angle, those would have a pitch of 5/12. To put this in a way that is easier to picture, imagine that the secondary rafters are 13 feet long, thus forming a 5-12-13 triangle. The would have a rise of 5 feet, a horizontal run of 12 feet, and an overall length of 13 feet. To keep the eaves of the roof level, the main rafters would, of course, have to drop the same 5 feet total at their end points. However, they would have to be longer than 13 feet. They would in fact be 13/cos(18) feet long. Thus, their pitch would actually be less than 5/12, and that's why the other formula is wrong. Josh this part, getting the miter angle, is way easier than the dot product solution. Consider 1/2 of one of the roof wedges. Vertically, it is a 12/5/13 right triangle, so the length from the peak to the eve is 13. Horizontally, it is an 18 degree right triangle, with the long leg of 12. Trig tables give us the other leg as 3.9. So the half wedge forming the roof is a right triangle with legs of 13 and 3.9, from which we use trig to get a miter angle of 16.7. It's the bevel angle that stumps me. I got an answer consistent with the one posted from the calculators, but without more digits of precision, I'm not very confident of mine. And unless one of you computer types want to "program" this solution in Excel gasp!, I'm not able to play around with the precision. My solution still causes brain strain when I try to reconstruct it! g -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#29
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Compound miter brainteaser
alexy wrote:
And unless one of you computer types want to "program" this solution in Excel gasp!, I'm not able to play around with the precision. My solution still causes brain strain when I try to reconstruct it! g Never mind. I created a spreadsheet solution that duplicates the web page cited here, and in so doing got my brain round the bevel angle. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#30
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Compound miter brainteaser
"Josh" wrote in message oups.com... Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? If the roof was flat (zero pitch) like a ten-sided deck, the miter would have been 36 degrees from perpendicular, Wouldn't it be 18 degrees? |
#31
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Compound miter brainteaser
"alexy" wrote in message ... "Josh" wrote: Not for those who don't remember their trigonometry: My father recently built a gazebo. Just for fun, he did it with ten sides, rather than the traditional six or eight. He gave the roof a 5/12 pitch (22.6 degree angle from horizontal). Thus, the roof was comprised of 10 triangular wedges. He sheathed the roof with planks forming concentric ten-sided rings around the center. At what angles did he have to miter the planks to get them to fit perfectly? Answer 1: 18 degrees, after first building a jig to hold the planks at a 5:12 slope in his chop saw, RAS, etc. g Answer 2: 16.7 degree miter and 7.1 degree bevel. If that's not right, then I'll have to solve it on something bigger than a post-it, and write out my steps more carefully! For what it's worth, I modeled it up in SolidWorks and I got a 16.57 degree miter with a 7.12 degree bevel. So, I think you're pretty close alexy. Relz |
#32
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Compound miter brainteaser
I'll bet you made the same mistake in SolidWorks that I alluded to in
an earlier post. If you were to start at the peak of the roof and draw 10 lines extending radially outward with a downward pitch of 5/12, you'd come up with angles of 16.57 and 7.12, as you did. However, for a 5/12 slope going straight down the roof (i.e. along a path bisecting two adjacent lines of the ten sided "starfish"), the slope along the ridges would not be 5/12; it would be (5/12)*cos(18). If you were to implement that slope in SolidWorks or Inventor or any other 3D modeling program, you should get the same answers as those given by DJ's algorithm (16.7 miter, and 6.8 bevel). Josh |
#33
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Compound miter brainteaser
"Josh" wrote in message oups.com... I'll bet you made the same mistake in SolidWorks that I alluded to in an earlier post. If you were to start at the peak of the roof and draw 10 lines extending radially outward with a downward pitch of 5/12, you'd come up with angles of 16.57 and 7.12, as you did. However, for a 5/12 slope going straight down the roof (i.e. along a path bisecting two adjacent lines of the ten sided "starfish"), the slope along the ridges would not be 5/12; it would be (5/12)*cos(18). If you were to implement that slope in SolidWorks or Inventor or any other 3D modeling program, you should get the same answers as those given by DJ's algorithm (16.7 miter, and 6.8 bevel). Josh I guess I don't understand what you're saying. Could you or someone please explain it to me? I've modeled two sections of a ten-sided roof with a 5/12 pitch. I've modeled two 1 x 4 "boards" that would join up on the theoretical middle of the rafters. For the boards to come together in both miter and bevel, the CAD software is still telling me that the miter should be 16.57358103 degrees and the bevel should be 7.12327393 degrees. Sorry, but I've learned to trust my CAD software and either you are all wrong, or more likely, I've put something into my program that's wrong. Also, if anyone has SolidWorks, I am willing to email you my file and maybe you could tell me what I'm doing wrong. Relz |
#34
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Compound miter brainteaser
"Relz" wrote:
"Josh" wrote in message roups.com... I'll bet you made the same mistake in SolidWorks that I alluded to in an earlier post. If you were to start at the peak of the roof and draw 10 lines extending radially outward with a downward pitch of 5/12, you'd come up with angles of 16.57 and 7.12, as you did. However, for a 5/12 slope going straight down the roof (i.e. along a path bisecting two adjacent lines of the ten sided "starfish"), the slope along the ridges would not be 5/12; it would be (5/12)*cos(18). If you were to implement that slope in SolidWorks or Inventor or any other 3D modeling program, you should get the same answers as those given by DJ's algorithm (16.7 miter, and 6.8 bevel). Josh I guess I don't understand what you're saying. Could you or someone please explain it to me? I'll try different terminology. I thin what you probably modeled has hip rafters at 5:12 rather than common rafters at 5:12. I don't have solidworks, but have another solid modeler. I'll try it and see what I get. I've modeled two sections of a ten-sided roof with a 5/12 pitch. I've modeled two 1 x 4 "boards" that would join up on the theoretical middle of the rafters. For the boards to come together in both miter and bevel, the CAD software is still telling me that the miter should be 16.57358103 degrees and the bevel should be 7.12327393 degrees. Sorry, but I've learned to trust my CAD software and either you are all wrong, or more likely, I've put something into my program that's wrong. Also, if anyone has SolidWorks, I am willing to email you my file and maybe you could tell me what I'm doing wrong. Relz -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#35
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Compound miter brainteaser
alexy wrote:
"Relz" wrote: I guess I don't understand what you're saying. Could you or someone please explain it to me? I'll try different terminology. I thin what you probably modeled has hip rafters at 5:12 rather than common rafters at 5:12. I don't have solidworks, but have another solid modeler. I'll try it and see what I get. I've modeled two sections of a ten-sided roof with a 5/12 pitch. I've modeled two 1 x 4 "boards" that would join up on the theoretical middle of the rafters. For the boards to come together in both miter and bevel, the CAD software is still telling me that the miter should be 16.57358103 degrees and the bevel should be 7.12327393 degrees. Actually, before I model, let me explain what I'd do, and see if what you did is equivalent. Start with a plan view of 1/20 of the gazebo: a right triangle with (for convenience) a long leg of length 12. Extrude this at 90 degrees, to a height greater than 5, leaving you with a triangular prism. on the surface defined by the long leg (not hypotenuse) of the first right triangle: draw another right triangle consisting of the top of this surface (12 units long), down one side 5 units, and the hypotenuse. Extrude cut (or whatever it is called in solidworks) with this triangle to cut away the top of the prism. What you have left is a solid that if reflected on the short side and repeated 10 times at 36 degrees would be the gazebo (but don't do that). Look at the shape of the "roof" piece to get the miter cut. Now define a plane perpendicular to the roof, and to the edge of the roof that is the hypotenuse in the plan view. Look at the angle on this plane of the intersection of the roof and the side. This is the bevel cut. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#36
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Compound miter brainteaser
If you can email me a .SAT file I can check it out. Given that you get
the exact angles that I get with my dot-product solution above, It seems like you must be defining the hip rafters at 5/12 pitch, which is slightly off from their actual pitch, but then again, maybe we're all wrong ;-) If you picture the roof comprised of 10 triangular wedges, the two long sides of each triangle would be the hips of the roof, and the short side would be the eave. Now bisect this triangle into two right triangles by drawing a line from the point of the roof to the center of the eve (If the triangle was a Christmas tree, you'd be drawing the trunk). This new line should have a pitch of exactly 5/12 (i.e. should form a 22.62 degree angle from horizontal). Meanwhile, the hip rafters should form an angle of 21.62 degrees from horizontal because they are slightly longer than the common rafter you just drew, but they rise the same distance. In other words, they're slope (rise/run) is less because "rise" is the same, but "run" is longer. If you measure the angle from your miter joint to horizontal (which should be the same as the pitch of the hip rafters), does it come out to 22.62 degrees or 21.62? |
#37
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Compound miter brainteaser
"Josh" wrote in message oups.com... If you can email me a .SAT file I can check it out. Given that you get the exact angles that I get with my dot-product solution above, It seems like you must be defining the hip rafters at 5/12 pitch, which is slightly off from their actual pitch, but then again, maybe we're all wrong ;-) If you picture the roof comprised of 10 triangular wedges, the two long sides of each triangle would be the hips of the roof, and the short side would be the eave. Now bisect this triangle into two right triangles by drawing a line from the point of the roof to the center of the eve (If the triangle was a Christmas tree, you'd be drawing the trunk). This new line should have a pitch of exactly 5/12 (i.e. should form a 22.62 degree angle from horizontal). Meanwhile, the hip rafters should form an angle of 21.62 degrees from horizontal because they are slightly longer than the common rafter you just drew, but they rise the same distance. In other words, they're slope (rise/run) is less because "rise" is the same, but "run" is longer. If you measure the angle from your miter joint to horizontal (which should be the same as the pitch of the hip rafters), does it come out to 22.62 degrees or 21.62? Okay, I think I know where I went wrong. I was modeling the 5/12 pitch on the rafter, not the roof (your tree trunk). I remodeled the senario as you described and I'm getting a hip rafter angle of 21.62, just as you described. And, now I am getting a bevel degree of 6.83 which is what everyone else was getting. Thank you Josh and alexy for helping me see where I went wrong. This information will come in handy when I go to build my gazebo. I just may do it ten sided! :-) Relz |
#38
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Compound miter brainteaser
Nice. I put a couple of pictures of my father's gazebo online.
They're not very good, but I think I have some better ones at home that I might upload. http://www.geocities.com/jcaron2/gazebo1.jpg http://www.geocities.com/jcaron2/gazebo2.jpg Josh |
#39
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Compound miter brainteaser
In article .com,
says... Not for those who don't remember their trigonometry: What would the formula be for N sides with a roof pitch of A degrees? an interesting problem. it is solved by noting that the rafters, when viewed from above, look like an outside corner crown moulding (c.m.) problem. the pitch of the rafters (5:12 = 22.6degrees) becomes the c.m. angle, while the inclusion angle becomes (180-36)/2 = 72degrees (again, since it's an outside corner crown moulding). in this specific case the result is: miter = pi/2 - [atan (cos(22.6) / tan (72)] = 16.7deg bevel = atan [cot(pi/2 - 22.6) * sin(pi/2 - 16.7) ] = 6.8degrees for those who want to visualize what's going on or who aren't necessarily good at trig, check out my tutorial at http://users.adelphia.net/~kimnach/woodworking/compoundangle.htm -- regards, greg http://users.adelphia.net/~kimnach |
#40
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Compound miter brainteaser
Surely this a simple problem. Akin to cutting crown molding, where
the number of sides are plugged into the 360 and divided by two to get the miter, and the spring is replicated on the bed of the cutting table, or mathed out to be cut on the flat according to existing tables. The more interesting concern is that the butting joints be undercut, so as to avoid expansion under the inevitable expansion under moisture absorbtion, which would destroy the geometry quickly. This is a carpenter's problem, rather than one for the mathematicians, and a decent carpenter would say to allow for a sixteenth between the joints, along with the underbevel, so that the whole structure has a decent chance of survival. (caveat - it is often assumed that there is no expansion along the lateral line but in this situation it must be alllowed for by creating a relief space and an undercut at the joints) |
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