*
Anyone? Looking for a math solution to
cutting groin arches.Are the four pieces that meet actually a triangle shape then bent?Picture at
www.ucr.edu/h-gig/maps/vault7.gif (e-mail [email protected])
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Scott,
The rectangle, ABCD, represents the PLAN VIEW for the "groin arches" as you refer to them. It you were framing a split- pitch hip roof ( a diamond hip roof ), the PLAN VIEW would be exactly the same, in which case, cutting triangular pieces would work.
The protusions that you see in the web site drawing are the actual shapes that the plywood would have to be cut to fit these "groin arches". Cutting triangles will not work.
I regret to say that I haven't taken the time to work out mathematical formulas for these curves, as I have not been confronted with this situation before, but, if you can give me a little time, I will look into this interesting problem and post a solution for you.
In the meantime, there are others here at Breaktime that may have experience with this type of detail, and you may find solutions forthcoming before mine.
How much time do you have before making these cuts?
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Excellent. Thankyou. I possibly could cut mine with this info, because the arch widths are equal.The different sized openings seems to throw me.
The spring lines are the starting and stopping point of the curve? The top height will always match it's opposing arch? The bottom spring line will be different if width is different? In our diagram does n always = x ? Are the protrusions in the drawing the same ellipse with axis swapped? How do I determine y's value ? I definately don't understand the green formula.
One final large note - I had thought the protrusions to be the shape to cut for the interior ceilings. Their not are they? Their merely the profile of the arch looking at it ? If so , sounds much more difficultryone.
It still seems like the ceiling pieces would be triangle based. A point on each side of the bottom of the arch and the "o" or top . If bending a piece c - d would be a line. C-O and d-o would either be straight or curve based? Thanks again everyone. Scott Walker
*Ken thankyou for your response. I have a couple months before this is needed.To Ken and others I apologize for any improper placement of info or replies, I'am not a computer man.However the info is worth it.I will further reply under Joe's diagram, as I,m still confused as to the unequalness of the two arches/ellipses.Thanks again Scott Walker
*
Scott,
If you want to build your confidence you might get a couple of pieces of 1/8 plywood and doing a scale model of what you want.
Rick Tuk
*Scott or Joe,Could you identify the name of the book or magazine in which you found this drawing? I'd like to read more on this subject, if it exists.
*
Ken I found illustrations by searching under groin arch or cross vault. There was very little info with this. A couple of replies posted (I don't know where) talk of cutting sheetrock and elliptical hips. This is the shape I'm hoping to define.I'm sure we now know what a groin arch is but how could we cut the interior pieces.
Thanks Scott
*
Joe, When I look at your drawing I am reminded of the great cathedrals in Europe.Am I the only one who is amazed that those guys built such tremendous creations without computers,with plumb bobs but not spirit levels,with squares but not framing squares and tables and certainly no speed squares andd the accompanying instruction booklet?Flying butresses,gargoyles,Hammer beams etc.A lot of cool medieval buildings.
I wonder if I.M Pei's work will stand up so well?
Good Luck,Stephen
*Scott,In your upcoming project, do you have a single barrel ceiling intersecting another barrel ceiling at right angles, or are there several barrel ceilings intersecting such as shown in the web drawing by John Fitchen?Also, what is the diameter(s) for the barrel ceilings that you will be constructing?
*Stephen Ken Follett wrote a book "Pillars of The Earth". A novel about cathedral building in the 12th century. Was a pretty good read.
*Joe,I own a copy of the book you refer to by George Collings, but it doesn't seem to address the question that I would ask.We all know what a "barrel vault", or a "barrel ceiling" is. It's just one half of a cylinder, what you would get if you cut a tin can in half lengthwise, so to speak.Now, if you take 3 of these half cylinders and place them adjacent to each other and then let them be intersected by a fourth half cylinder, running at right angles to the first three, you will get a so called "groin vault" result, at least as I see it.So, basically, a groin vault appears to be nothing more than several "barrel ceilings" intersecting each other. How do you see it?
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Ken - Yes I have a single arch intersecting another at right angles. Both are equal widths(thankfully) of 5'6.The heights are unspecified, as are so many times when someone else wouldn't commit.While the plan calls for a groin arch, it is possible a intersecting barrel is what they intended. Matching other single arches in the house. Seems Joe may know how to solve this?
Thanks for the thought - Scott
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Scott and Joe,
I did a search and found Mr. Fitchen's Gothic Cathedral details at this site
I still feel that a "groin vault" is nothing more than several intersecting "barrel vaults" in a certain pattern. Notice that after the words groin vault that in parenthesis are the words "cross vault", which implies that this may be the case; a descriptive word for a pattern of intersecting barrels.
*Scott,I believe that as time goes by that you will find yourself constructing two intersecting barrel vaults ( half cylinders) at right angles to one another.If this proves to be true, then the lines of intersection will be elliptical ( elliptical valleys, as seen from above). Since the diameter of each vault is 66", this will also be the length of the minor axis of the ellipse. The major axis would simply be the diagonal of a square where each side of the square is 66", which would be 66" times the square root of 2,66" x 1.4142 = 93.338" or 93 5/16"The two foci of the ellipse would be at 33" from the center of the ellipse, on the major axis.This makes it possible to draw the "half ellipse" on a single sheet of plywood. I'm not sure how much experience you've had with this Scott, so I'll give a description of how to draw it.1)Snap a line running with the length of the plywood a few inches in from one of the long sides of the sheet of plywood.2) mark the center point of this line at 48" in from either side. This point will be the center of the ellipse, and the major axis will lie on this line.3) measure out 33" in each direction from the center point and mark the two foci of the ellipse on the major axis. Tack a small finish nail at each of these foci points.4) draw a line that is square to the major axis beginning at the center ( as if you would do to snap a line to cut a sheet of plywood in half)Measure up 33" from the center point of the ellipse and tack another finish nail on this line. (this line is one half of the minor axis)5) you're now ready to draw the ellipse. If you use string, you'll get a fair result, but if you want a more accurate drawing, buy a length of picture frame wire at the local hardware store. It's excellent for this purpose as it doesn't stretch and is very flexible. Try to buy a piece about 10 feet long or more.6) loop one end of the wire around one of the two foci nails, pull it over the top of the nail on the square line and then back down to the second foci and loop it off again.7) remove the nail on the square line and replace it with a sharpened pencil. Keeping constant pressure against the wire, move the pencil from side to side to draw the half ellipse When cutting the ellipse, place a second sheet of plywood directly below the first so that you can cut both half ellipses at the same time. After you've made the cut, place one of the half ellipses up on top of the walls diagonally. If the framing is accurate, it should fit just right from corner to opposite corner. Now cut the second half ellipse exactly in half on the square line that is drawn on it. Remove an extra 1/4" from each side for the thickness of the first sheet of 1/2" plywood, and attach each half to the first piece that is already in place, (much in the same way as you would put up the first four common rafters when building a hip roof on a square).You now have the elliptical valleys of intersection defined, and you can frame to them. You can add support by nailing a beam to the part of the plywood that lies above the ceiling and bracing it properly.I would like you to know that I have never actually constructed one of these things, but I believe that this method would serve you well. Perhaps others with more experience with this will post and offer improvements and perhaps completely different approaches. I look forward to hearing from them, as I'm sure you do also.Good luck...Ken
*
All those tools aside, I'm awed to think they did all of that without electricity and the kinds of power tools that we have. Hammers and chisels to detail marble for hundreds of columns and thousands of blocks? No way...
-Ben
View Image
...and remember,
Strip Mining Prevents Forest Fires.
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Joe,
Your points are well taken, and , another nice CAD drawing by the way.
The only thing I would disagree with is the statement at the end of your post that states that the green lines that are the minor axis of the ellipses is the major axis for my ellipse. Clearly, I stated that the major axis would be equal in length to the DIAGONAL of the square, each side of the square being 66". In your diagram, that would be represented by the red lines that are the diagonals of the square base.
As you point out however, the method that I described would only work for two barrel ceilings of equal radii ( semicircular ) that intersect at right angles.
The diagram that you present creates some very interesting possibilities. The "ceilings" that are going to intersect at right angles, are not semi-circular, but rather, elliptical, as represented by the blue lines. When these elliptical surfaces intersect, the result should still be elliptical valleys, ( the green ellipses ), but in the special case where the side of the base square times the square root of 2 ( which would be either of the red diagonals of the base square) is equal in length to the major axis of the blue ellipses, the resulting intersecting valleys would actually be CIRCULAR, the diameter of the circles equal to the diagonal of the base square.
Suppose Scott wanted to make his "ceilings" elliptical as as shown by the blue ellipses in the CAD drawing. Since his minor axis ( the green lines of the base square) for the blue ellipses will be 66", then the red diagonals of the base square will be = 66" times the square root of 2, or 93 5/16". If he now were to make the major axis of the blue ellipses equal to 93 5/16", then the intersecting valleys would be circles, which would be easy to frame)
In all other cases, the resulting intersections would be elliptical, and depend on the relationship ( the ratio ) between the major axis and the minor axis of the blue ellipses. ( which I assume would be given in the plans)
If the red diagonal of the base square is larger that the full major axis of the blue ellipse, then it will be the major axis of the green ellipse. If however, it is smaller than the major axis of the blue ellipse, then it will become the minor axis of the green ellipse, and the major axis of the blue ellipse will become the major axis of the green ellipse. ( that was a mouthful )
I want it to be clear in the above discussion that the terms major axis and minor axis, by definition, always refer to the entire length of the axis. If the measurement of either axis is taken from the centerpoint of an ellipse, the result will be the length of the semi-major or semi-minor axis. ( half of the axis)
*Joe,Thanks for the reply. I apologize for the apparent breakdown in communication on my part. I, for some strange reason, thought when you said "in green" you were refering to the green lines of the base triangle, and not the green ellipses. Sorry about that. ( Don't you have any more colors? Otherwise, I would like you to read the rest of my post now that I have admitted to misinterpretation of yours, and comment. I think that you'll find it interesting that the minor axis that you refer to in your CAD drawing, will not always be the minor axis. It may just as well turn out to be the major axis, depending on the ratio of the major axis to the minor axis in the blue ellipses.
*Joe,By "easier to frame", I was suggesting that once it was determined that the intersecting cross vaults would be elliptical(as in your CAD drawing) rather than "barrel" (semi-circular), a good choice for these ellipses might be one in which the ratio of the major axis to the minor axis would result in intersecting valleys that would be circular. If the length of each of the "green lines" that form the square base in your CAD drawing were designated by "x", then the diagonal of the square ( your red line diagonals) would be the length "x" times the square root of 2. If you chose the major axis of the "blue ellipses" to be of this length, then the resulting valley intersections would be circular. Swinging a circle on a sheet of plywood to define the intersecting valleys, would surely be "easier" than creating yet another elliptical shape to define them. Of course, as you point out, you would still have to construct the ellipses to define the vault, but, as you show in your CAD rendering, this would always be necessary whenever the ceiling vault were to be elliptical instead of barrel.
*Ahhh...the good old days of indentured servitude...
*Joe, I agree that the elliptical shapes would have to "look nice" as you put it. As a matter of fact, the elliptical shapes that would result in circular valley intersections that we have been discussing are actually quite pleasing to the eye, the ratio of the major axis to the minor axis being 1.4142, not far off from the "golden ratio" of 1.618034, which is by some peoples judgement, the most pleasing ratio to the eye.I'm not suggesting by any means that all such "cross vaults" should be constructed in this manner. I was simply replying to your suggestion in an earlier post in this thread that perhaps I might come up with a general approach to this subject which would include "elliptical cross vaults" instead of just the typical "barrel cross vault" intersection.Normally, we are faced with the situation of two semi-circular vaults intersecting each other at right angles to form elliptical valleys (or hips, depending on the configuration). I thought it might be interesting to you and to others who are reading this thread, that it is also possible for intersecting elliptical vaults to intersect in circular valley intersections, the opposite situation. It just so happens, that if one were required to construct such a detail, the ratios that I speak of would be an acceptable choice, and that being the case, would be a good choice since the circular valley intersections would be easier to construct, rather than having to include yet another elliptical shape.
*Joe,I don't have much to add at this point except that I noticed when looking back at that nice CAD drawing that you made of the "elliptical cross vault" that it is very close to the situation I was describing in which the intersections of the elliptical surfaces would turn out to be circular.The blue ellipses that define the vaults are very much elliptical, but notice that green ellipses are very close to being circular. If the height of the blue ellipses were reduced just slightly, the green ellipses would be perfectly circular. As I pointed out, this would occur when the length of the major axis of the blue ellipses is 1.4142 times the length of its minor axis.The only point I am making here is that your CAD drawing results in a shape that is very pleasing to the eye, and it just so happens to be very close to the situation that I described.
*
Ken And Joe I have been paying attention, but was lost for some time. Maybe still.Let me try -
an intersecting barrel vault-
- arches always equal widths and equal spring lines
- Elliptical "hip" arches always.
- minor axis equals the width of the arch
- major axis equal to diagonal of square
(which could never be rectangle because width
are equal?)
-this is not a template, it is the piece that would fit the ceiling (like drywall)?
a groin vault -
- arches elliptical
- elliptical hip arches, unless major axis =
minor axis x square root 2
- a circular hip would have a diameter of ??
I am familiar with cutting circles and ellipses, but for a scale model is there another way to lay out an ellipse w/o using a mini string ? Thanks again(feeling dumb, but always like learning) Scott
*
Scott,
Glad to see you're still with us.
I know of no easier way to quickly swing an ellipse than the one that I described earlier in this post. I often do them in this manner and get very accurate results, even with scale models.
I'm curious to find out if you want to build barrel ( semi-circular) vaults, or semi-elliptical vaults. Let us know, will you?
Remember, if you decide to make them semi-circular, the intersetions will always be an ellipse with the following properties:
1) the minor axis will always be equal to the diameter of the circle ( for example 66" )
2) the major axis will always be equal to the length of the minor axis times the square root of 2 ( for example 66" x 1.4142 = 93 5/16" )
3) the foci points of the ellipse will always be equal to 1/2 of the minor axis ( for example 1/2 x 66" = 33" )
4) the length of the string or wire to draw the ellipse will always be equal to the length of the major axis ( for example 93 5/16" )
*Nice math, folks, and the drawings are great.Scott, from your last post I see we haven't answed your original question, which is to paraphrase (stop me if I'm wrong here), what shape are the ceiling panels under the groin--that is how does one lay out and cut the drywall (or whatever) to fit perfectly once the groin is framed?My short answer is it won't be a triangle but rather a piece with two curved sides and one straight side. The shape of the curves is determined by the shape of the ceiling and in my opinion would be best determined by trial and error/actual measurement in place, both because once you start wrestling the material in place you are likely to find the mathematical formula a bit off, and also because I don't know the formula (grin).Another possible clarification to this discussion. I always thought a groin was the intersection between two vaults, no matter the shape of the vaults. Scott, although you gave us the width of the vaults, you haven't told us the shape of them, which could be half-circles (barrel vaults) or half ellipses, or some other shape like a gothic arch which if I remember is made of two arcs which make a pointy intersection at the top, or a false ellipse or some other such curvy shape.Whichever shape the arch is, the groin "hips" in your (equal width vaults) situation will have the same height but a lenghth equal to the width of the vaults times the sqare root of two. The other points on the curve follow from there but for my money the vaults had better be a half circle or half true ellipse or the math will be so complicated on the hips you'd be better off using string lines and patience to figure them out.One other hint for actual construction--I would consider furring the vaults with 1x material to give more surface to pull ceiling material up to.There is another way to get ellipses and that's with a set of tracks mounted in a cross and a bar that... oh, this is too complicated to describe verbally. Write back for more info--someone has been advertising a router jig of this type in Fine Woodworking lately and I could look it up for you if you'd like.Good luck. I envy you good fortune in landing a project like that.Doug
*Doug,Thanks for joining the discussion.I'm working on the formulas ( if formulas actually exist ) for the cuts that would have to be made for the drywall on the underside of the "groin". I must say, at this point, it isn't likely that the formulas will be helpful ( if they exist ), as I have already ruled out straight lines and circles for these unusual curves ( cuts ). I am at the point that ellipses also look doubtful. This leaves other possibilities, such as parabolas, hyperbolas, or possibly even sine or cosine curves, which would be ridiculous to ask a framer to consider using. Trial and error would be a faster method if this were the case.Stay tuned. I intend to get to the bottom of this one, even if the information turns out to be impractical. In the meantime, I'll be following this thread, and look forward to additional posts from you and people such as Joe who can add to this topic.
*
I'm blessed to be surrounded by greatness and possibly gluttons (for punishment). It has come to be more about principle than specifics. I agree.
While either yourself or Joe (or others) could tell me much easier than teach me, which may be necessary yet, I'm hoping to learn.
Doubtful that I will understand the theory, maybe I can resolve "theorems " that will let me fish for a life time so to speak. Several specifics have been defined nicely.
I haven't had time but will soon start my mini Groin Town(out of const. papers) .
I have believed the shapes we defined as templates to fit against (which defines the shape)
Which the upper portions of these cuts could be the hips themselves.
But the framing in between the hips will not be straight either?(anyone know the diameter of Jack's head)
Good luck guys, I will keep checking, but bola's are way past me.
See Ya Scott W.
*If you are framing these things, I guess I think the formulas are the place to start (s0rry). If you just have to rock or sheath them and want to figure out the shape of the skin, try this.measure from the barrel entrance to the valley at the lowest spot. Then measure on inch higher, and keep going until you have 50 (or whaterer) measurements. Lay these out on your material and cut.The same procedure works for an intersecting area, start in the middle where to valley's come together and measure the width. Keep measuring as you move your tape up every inch and lay it out. It's time consuming but sounds better than construction paper.
*
Anyone? Looking for a math solution to
cutting groin arches.Are the four pieces that meet actually a triangle shape then bent?Picture at
http://www.ucr.edu/h-gig/maps/vault7.gif (e-mail [email protected])