PDA

View Full Version : Figuring plane change degrees


aeviaanah
05-23-2011, 07:25 PM
Typically when laying out transitions or plane changing fittings (such as a twist) I would layout as usual and form the plane change until it hit somewhere close. I have asked lots of sheetmetal journeyman if they know how to figure the actual degree which to brake the plane change. Can anyone explain to me how to figure this using layout without computers.

I can figure this degree using computers...I draft the 3d fittings point A to point B, i then draw in my plane change brake and measure the angle. I use google sketchup for this. I then export and form after burning on plasma table.

As I am interested in the way the ol timers used to do it...Id like to know how to figure this degree without estimating. Anyone can form a 24 gauge square to round or twist and bend it around until it lives where its supposed to...im talking doing it mathematically without farting around. Ive attached three images of a sample fitting....Fill free to plug in dimensions or whatnot...this fitting is a 24 x 24 bottom, 12 x 12 top, 18 tall and twists 45 degrees....i didnt provide the offset numbers.

Feel free to explain it with whatever numbers...

Attached is a plan view, elevation and an isometric. Thanks!

Bud
05-24-2011, 04:47 AM
To find the degree of the angle you would have to create the drawing looking at the angle or end view perpendicular to it's plane.

This doesn't mean that the end of the sheet as to be on a plane perpendicular to the bend line because that won't matter if you're just trying to get an actual angle of the bend line.

If you need help with that a good reference book would be something like instructions on descriptive geometry where they can show you how to create views that are perpendicular to the plane

roofermarc
05-24-2011, 07:46 AM
smwlocal24 might be able to word it correctly step by step, great explanatory skills as already demonstrated. My moneys on him. No offense Bud.

aeviaanah
05-24-2011, 09:02 AM
To find the degree of the angle you would have to create the drawing looking at the angle or end view perpendicular to it's plane.

This doesn't mean that the end of the sheet as to be on a plane perpendicular to the bend line because that won't matter if you're just trying to get an actual angle of the bend line.

If you need help with that a good reference book would be something like instructions on descriptive geometry where they can show you how to create views that are perpendicular to the plane

thanks bud, this is the only idea i could think of yesterday. im just not sure how to would construct the geometry. ill try to find a book...seen anything of the sort in the sheet metal shops library?



roofermarc, thanks ill track him down and see if he'll come chat about the topic. thanks!

Bud
05-24-2011, 10:50 AM
smwlocal24 might be able to word it correctly step by step, great explanatory skills as already demonstrated. My moneys on him. No offense Bud.

No offence taken/ let's see what he's got for us. Im thinking that might be a bit hard to explain without a drawing or basic understanding how to see things in its true length or true shape. Maybe im reading the question wrong?

Bud
05-24-2011, 10:55 AM
thanks bud, this is the only idea i could think of yesterday. im just not sure how to would construct the geometry. ill try to find a book...seen anything of the sort in the sheet metal shops library?



roofermarc, thanks ill track him down and see if he'll come chat about the topic. thanks!

I have a couple great books on the subject/ I can get you the isbn to them . Ill look in all the manuals I have ads well. Pictures always worked best for me.im sure there is a simple mathematical approach too.
Ill check back later.

smwlocal24
05-24-2011, 09:13 PM
First off, this is very intensive math for anyone who hasn't seen or heard of any of this stuff before, so it is very likely that I will have to clarify some portions of this method.

The best way to calculate an angle on complex geometry such as this is by using vectors to represent edges of your fitting. A vector is kind of like a line that has a starting point and an ending point where the starting point, for convenience, is usually thought of as being located at (0,0,0) in the XYZ cartesian coordinate system and the end point is located at some (x,y,z) within that same system. So to describe a vector you would only need to identify its ending point location like this <x,y,z>.

To find the angle between two intersecting planes you must:

1.) Identify two different vectors that lie within each plane based on what you know about each plane's geometry (make sure to label them so that you can keep them straight). You will have to set up your own right-handed XYZ coordinate system for these planes and vectors to be described in. If you are looking down the positive x-axis in a right handed system, then the positive y-axis is to your left and the positive z-axis is pointing up.
http://en.citizendium.org/images/thumb/e/e4/Right-hand-rule.jpg/350px-Right-hand-rule.jpg (http://en.citizendium.org/wiki/File:Right-hand-rule.jpg)
2.) Using the right-hand rule, take the cross product of the pair of vectors within the first plane and this will give you a vector that is normal(perpendicular in all directions) to the first plane's surface. Repeat this step for the second plane's vectors.

Say for plane 1 you have vectors A and B, where A=<x1,y1,z1>
and B=<x2,y2,z2>

The cross product between A and B would be:
A X B = < (y1*z2-z1*y2), -(x1*z2-z1*x2), (x1*y2-y1*x2) >


3.) Divide each of these normal vectors by their magnitudes(lengths).

The magnitude or length of A is equal to: sqrt(x1^2 + y1^2 + z1^2)

You would use this same procedure to find the magnitude of the normal vector formed by the cross product A X B and divide each of its components(x, y, and z components) by this magnitude and this will make your vector a normal unit vector meaning it has a length or magnitude of one.

4.) Take the dot product of your two normal unit vectors.

For vectors A and B, the dot product of A and B would be:
A dot B = x1*x2 + y1*y2 +z1*z2

5.) Take the inverse cosine of this dot product and the result is the angle that you will bend your metal at the intersection of those two planes.

I hope this is understandable, but I realize it might not be. There are a lot of different concepts that tie in together here and I'm trying to keep it somewhat short. Let me know what needs to be made more clear and I will get back to you as soon as I can.

smwlocal24
05-24-2011, 10:22 PM
Here is an example of the procedure I explained above. I solved for the angle formed by the intersection of the two planes in the lower left corner of the first picture posted. Check this result with your drafting software.

This method takes a very long time to explain, but is very efficient for a person that knows what they are doing.

roofermarc
05-25-2011, 05:54 AM
I'll stick to guessing at the degree of angle to fold the lines, but I told you that guy could explain it. A real student of layout. Thanks for sharing.

sharpscriber
05-25-2011, 02:27 PM
Hello aeviaanah (http://www.thesheetmetalshop.com/forum/member.php?u=3560),
You can develop that in one piece, right? I mean not for practicality but it can be done, right?
Well if the total of the entire bends is 360 deg. then in order to come up with the proper degree for each bend you can see where two of your vertical height from base to top points at the top twisted opening can go straight up perpendicular and the other two will finally meet. 4 bends will be symmetrical with themselves, negating out to 180 deg. when formed (bottom)
True lengths between the total lengths of one plane to another vs. the amount of difference from one another. That is: A true 90 deg. bent duct corner goes up so much but over none. But your slope on one plane goes up so much and over so much. The other one goes up so much and over a DIFFERENT amount. Since your height is the same this is easy. Two of your bends must equal a 90 deg. SSSSoooooooo you divide one into the other like a fraction and you can factor that into your 90. Say if one was 40% different than the other. THen your bend would be 40/100 out of 45/90 or 36 deg. The other one would be 60/100 out of 45/90 or 54 deg. Think about correlation and relationship with one another.
You can do different height configurations like this and just a square and a bench rule can give you the combined true lengths of a plane to figure out the amount of bend. If you want other than a 90 deg. then plug in those numbers instead of the 90 or change the 100%.
Might just try a couple of those just to make some aluminum light shades outside against the corner of walls. Looks deviously challenging. Loved studying this one. I still like the others' replies on this. Local 24 likes to stretch his brain on this stuff. Like to see that on paper done by hand. I would have sat next to him in college. haha:cool:

smwlocal24
05-25-2011, 04:06 PM
I misrepresented plane A in my example calculation posted above. The vector V that I used is not in plane A. I have recalculated the bend angle with a corrected vector V and have attached those hand calculations.

sharpscriber
05-25-2011, 05:51 PM
So sitting next to you in school probably would have failed me as I can't get into that trig so much as I used to. You mean I missed one of your mistakes? Should have known better than to go over your work without a calculator. I use calculations for some things but angles and length layout I like rule and simpler math. That way you can input your bend tolerances and thicknesses when using z dimensional material where it is desired. Thicknesses will change the bend line location.

aeviaanah
05-26-2011, 09:00 PM
First off, this is very intensive math for anyone who hasn't seen or heard of any of this stuff before, so it is very likely that I will have to clarify some portions of this method.

The best way to calculate an angle on complex geometry such as this is by using vectors to represent edges of your fitting. A vector is kind of like a line that has a starting point and an ending point where the starting point, for convenience, is usually thought of as being located at (0,0,0) in the XYZ cartesian coordinate system and the end point is located at some (x,y,z) within that same system. So to describe a vector you would only need to identify its ending point location like this <x,y,z>.

To find the angle between two intersecting planes you must:

1.) Identify two different vectors that lie within each plane based on what you know about each plane's geometry (make sure to label them so that you can keep them straight). You will have to set up your own right-handed XYZ coordinate system for these planes and vectors to be described in. If you are looking down the positive x-axis in a right handed system, then the positive y-axis is to your left and the positive z-axis is pointing up.
http://en.citizendium.org/images/thumb/e/e4/Right-hand-rule.jpg/350px-Right-hand-rule.jpg (http://en.citizendium.org/wiki/File:Right-hand-rule.jpg)
2.) Using the right-hand rule, take the cross product of the pair of vectors within the first plane and this will give you a vector that is normal(perpendicular in all directions) to the first plane's surface. Repeat this step for the second plane's vectors.

Say for plane 1 you have vectors A and B, where A=<x1,y1,z1>
and B=<x2,y2,z2>

The cross product between A and B would be:
A X B = < (y1*z2-z1*y2), -(x1*z2-z1*x2), (x1*y2-y1*x2) >


3.) Divide each of these normal vectors by their magnitudes(lengths).

The magnitude or length of A is equal to: sqrt(x1^2 + y1^2 + z1^2)

You would use this same procedure to find the magnitude of the normal vector formed by the cross product A X B and divide each of its components(x, y, and z components) by this magnitude and this will make your vector a normal unit vector meaning it has a length or magnitude of one.

4.) Take the dot product of your two normal unit vectors.

For vectors A and B, the dot product of A and B would be:
A dot B = x1*x2 + y1*y2 +z1*z2

5.) Take the inverse cosine of this dot product and the result is the angle that you will bend your metal at the intersection of those two planes.

I hope this is understandable, but I realize it might not be. There are a lot of different concepts that tie in together here and I'm trying to keep it somewhat short. Let me know what needs to be made more clear and I will get back to you as soon as I can.
Man you are right about this....complicated. I wish I had the mathematical background to understand this. I will give it another try but i seem to get lost around step 2. I read the whole tutorial thinking i may be able to get passed step two but I cant seem to. This is the type of problem that stumps even the best sheetmetal workers out there. I need some hands on over the shoulder training for something like this. Which math courses would you recommend to getting started to understanding this method? This seems like a great way to find angles....if practiced and put into use how fast could you find all angles in the posted fitting?
Here is an example of the procedure I explained above. I solved for the angle formed by the intersection of the two planes in the lower left corner of the first picture posted. Check this result with your drafting software.

This method takes a very long time to explain, but is very efficient for a person that knows what they are doing.
Again, not sure how patient you are but i am really wanting to understand this...Unfortunately i only exported as an image and have no way of checking the actual angle. Its seems pretty stupid to do such a thing, i apologize. I appreciate taking the time to try to explain this and I am looking forward to understand it.

Hello aeviaanah (http://www.thesheetmetalshop.com/forum/member.php?u=3560),
You can develop that in one piece, right? I mean not for practicality but it can be done, right?
Well if the total of the entire bends is 360 deg. then in order to come up with the proper degree for each bend you can see where two of your vertical height from base to top points at the top twisted opening can go straight up perpendicular and the other two will finally meet. 4 bends will be symmetrical with themselves, negating out to 180 deg. when formed (bottom)
True lengths between the total lengths of one plane to another vs. the amount of difference from one another. That is: A true 90 deg. bent duct corner goes up so much but over none. But your slope on one plane goes up so much and over so much. The other one goes up so much and over a DIFFERENT amount. Since your height is the same this is easy. Two of your bends must equal a 90 deg. SSSSoooooooo you divide one into the other like a fraction and you can factor that into your 90. Say if one was 40% different than the other. THen your bend would be 40/100 out of 45/90 or 36 deg. The other one would be 60/100 out of 45/90 or 54 deg. Think about correlation and relationship with one another.
You can do different height configurations like this and just a square and a bench rule can give you the combined true lengths of a plane to figure out the amount of bend. If you want other than a 90 deg. then plug in those numbers instead of the 90 or change the 100%.
Might just try a couple of those just to make some aluminum light shades outside against the corner of walls. Looks deviously challenging. Loved studying this one. I still like the others' replies on this. Local 24 likes to stretch his brain on this stuff. Like to see that on paper done by hand. I would have sat next to him in college. haha:cool:
This seems like an excellent way to explain such a thing. Again, i am lost tho. I feel this is a method i can understand and would like some further explanation...a few questions..

how would you figure the difference in percentages? i am not quite grasping this. can you provide a hand drawn explanation such as local24 did? feel free to construct your own fitting...one that may be easier to understand. thanks for the help...much appreciated!

I have a couple great books on the subject/ I can get you the isbn to them . Ill look in all the manuals I have ads well. Pictures always worked best for me.im sure there is a simple mathematical approach too.
Ill check back later.
hey bud thanks for the help... not sure what a isbn is. let me know some titles and ill get to lookin around for them. thankyou!

smwlocal24
05-27-2011, 03:18 PM
Which math courses would you recommend to getting started to understanding this method? This seems like a great way to find angles....if practiced and put into use how fast could you find all angles in the posted fitting?

The use of vectors, cross products, dot products, and the angle between two vectors is usually taught somewhere in a calculus sequence. Although there is no calculus needed to use this method, it may have been used to derive/prove the relationships/usefulness of cross products and such. You do not need calculus to understand this method; I recommend researching on the web the cross product of vectors, vectors, the dot product of vectors, and right handed coordinate systems.

For that particular fitting it could take some time to figure all the angles involved, and making a mistake is easy to do as I've already demonstrated. I would guess that it takes about 2-5 minutes to calculate one angle by the time you choose your coordinate system axes and identify arbitrary vectors included in two intersecting planes and then run the numbers. Most of this time is taken up setting up the problem and making sure you're getting it right; crunching the numbers doesn't take that long. With a fitting like this you would save some time because each plane intersects two other planes making the setting up of your next problem only half the work. To make this more economical, it would be best to make the fitting more symmetrical if possible which would lead to solving for much fewer angles. I use this method quite often but it usually is only for a few angles. I work in the architectural side of the trade, so I normally only use this for finding hip/valley angles of various slopes whether it be a hip cap, valley flashing, or a cricket above a chimney.

smwlocal24
05-27-2011, 03:37 PM
Well if the total of the entire bends is 360 deg.

I might not be interpreting this correctly but I think this is incorrect. The total amount of degrees bent to go around this fitting would not necessarilly equal 360 degrees. The reason I am thinking this way is that in some other known cases of layout they do not add up to 360. For example a hip roof or a square-based pyramid maybe. Suppose you have a square-based pyramid with sides that are sloped 45 degrees from the base. The bend angle of the intersection of any two planes would be exactly 60 degrees. With four intersections or bends this would only be:

4 * 60 deg = 240 deg.

Maybe it is 360 deg in total bends for this particular fitting, but I doubt it. I will have to check by finding all the angles involved; this will allow me to better estimate the amount of time it takes to completely solve a fitting like this for aeviaanah.

sharpscriber
05-27-2011, 03:44 PM
Correct on the total amount of angled bends on a slope. That's why I took the differences in ratio to the triangulated length and depth of both planes. The resulting ratio is a percentage of the total 360. I'm much more comfortable doing it on the bench rather than trying to explain it in words. Always works for me. Then I set up my digital angle finder on the brake and compensate for the spring back after running a test bend.

Cauthen
05-27-2011, 03:47 PM
I'm sure its either too late or this has been said before, but here's the easiest way. Just count up the number of total brake lines from its flat pattern. Then divide that by 360.


a joint of square duct has four sides. 360/4=90

EDIT: So your fitting's answer is an easy, perfect 45 degrees. 360/8=45.

Think about a regular old hem. It has two sides. 360/2=180

smwlocal24
05-27-2011, 04:42 PM
Correct on the total amount of angled bends on a slope. That's why I took the differences in ratio to the triangulated length and depth of both planes. The resulting ratio is a percentage of the total 360. I'm much more comfortable doing it on the bench rather than trying to explain it in words. Always works for me. Then I set up my digital angle finder on the brake and compensate for the spring back after running a test bend.

So by your method, what angle would you bend the "hips" of a square-based pyramid whose sloped sides are 25 degrees from its base?

smwlocal24
05-27-2011, 04:55 PM
This seems like a great way to find angles....if practiced and put into use how fast could you find all angles in the posted fitting?

starting with the angle i solved for in my earlier post and working counter-clockwise the angles in degrees are 55.169, 38.515, 67.645, 55.169, 41.897, 47.869, 47.869, and 41.897. These add up to 396.03 degrees. It took me 35 minutes to do this with some distractions(wife, kids, etc.) and I didn't take time to notice that there was some symmetry to the fitting which would have saved a little time I think.

Sharpscriber, could you solve for one of these angles with your method to check and see if we obtain the same results?

smwlocal24
05-27-2011, 04:57 PM
I'm sure its either too late or this has been said before, but here's the easiest way. Just count up the number of total brake lines from its flat pattern. Then divide that by 360.


a joint of square duct has four sides. 360/4=90

EDIT: So your fitting's answer is an easy, perfect 45 degrees. 360/8=45.

Think about a regular old hem. It has two sides. 360/2=180


I don't think you understand the problem. What you've described is an octaganol straight duct which is completely different than this.

sharpscriber
05-27-2011, 04:59 PM
I'll try to explain it this way smwlocal24: Take your pyramid model with its base and put a square on the bend until the square fits properly on the corner. It will be at an angle. Of course a square duct with bends perpendicular to the base will be in plane with the base. The angle of the measuring square on the pyramid is not parallel with the base now but this approach is what I am getting to when I say take the ratio of the triangulated height and horizontal differences between the two sets of planes.
Remember that the first example given had 8 total bends to it. Altogether they will still give a fitting a 360 total return but your compensation for the planes being off perpendicularity is a ratio between this 360, number of bends and the relationship between one plane to another and the vertical offset of the plane. Same as you're doing in your trig calcs but I just use measuring tools of the bench. I would rather use a calculator to give the bends but not fun for some of us lazy folks. I can use circular radial line development also in this with squared angles in it to get me the bend degrees but that requires a protractor to measure the final layout lines without the inconvenience of going through an extra two steps in measuring them without one.

john_galt
05-27-2011, 05:51 PM
This is a thought provoking thread, good stuff. Keep going and I'll try too keep up.
I will do scale models or test bends if I can't confidentially figure it on my calculator.

sharpscriber
05-27-2011, 06:05 PM
I hung up my calculator after college. Ha! I have a scientific calculator that costs over $100 that I don't use anymore. I do need to divide stuff always and junk so I use a simple one for that.
Take the relationship between the distance from a base corner of a pyramid to the top corner's travel. Then take the plane one of the sides is on from the middle of the base to its travel at the top. There is a relationship between that, the number of bends in the piece, what the finished bends are, whether 360 or any other bends and that ratio. Hate to let fun go as I would like to get to the bench and do some stuff with it too but I am on bidding spree at my favorite online auction and stuff is ending soon almost back to back.

sharpscriber
05-27-2011, 06:18 PM
This is scaring me to no end. Here's a diagram with the corners and points circled. There is a relationship between the differences from one to another measurements and that will fit into a proper ratio of the difference of a 'midline' bend angle for say a 45 deg. for an 8 bend piece. It will be less on sloped panels or planes and more if the walls were overhanging.
This is an excellent way to understand what to divide into what and subtract from what as you start from doing one just like a square piece of duct with 0 divided by 0 which is no change from 90 deg. But it is 4 bends.
Just do a simple irregular pyramid and then measure these and you will see the corresponding bend angle fit exactly as the ratio difference of their relationships. Brain just blew a fuse. Gotta go.

aeviaanah
05-27-2011, 06:38 PM
This is scaring me to no end. Here's a diagram with the corners and points circled. There is a relationship between the differences from one to another measurements and that will fit into a proper ratio of the difference of a 'midline' bend angle for say a 45 deg. for an 8 bend piece. It will be less on sloped panels or planes and more if the walls were overhanging.
This is an excellent way to understand what to divide into what and subtract from what as you start from doing one just like a square piece of duct with 0 divided by 0 which is no change from 90 deg. But it is 4 bends.
Just do a simple irregular pyramid and then measure these and you will see the corresponding bend angle fit exactly as the ratio difference of their relationships. Brain just blew a fuse. Gotta go.
Can you walk me through this one more time? I'm still confused!! I need more or less a step by step explanation....this is getting interesting!



Anyone ever been to Railtown in Jamestown, CA? i rebuilt the jacket for the famous Sierra Number 3. Shes been in over 100 movies such as Back to the future 3, high noon, etc.

aeviaanah
05-27-2011, 06:41 PM
http://railtown1897.files.wordpress.com/2010/09/img_1364-s.jpg

smwlocal24
05-27-2011, 07:25 PM
Sharpscriber, what bend angle do you get for the bendline you have highlighted in your sample pic. This would clear up everything. I am confused about what values you are plugging in for a ratio, and what to do with that ratio when I get it.

sharpscriber
05-27-2011, 09:16 PM
When I get a chance to go to the shop to do one I can elaborate. Just think of worse case scenarios. One with perpendicular planes like a piece of duct and one flat piece of metal just a square piece, no bends. One will work out to 90 deg. bends and the other will work out to 0 deg. bends. All the rest in between is a relationship between the both.

smwlocal24
05-27-2011, 10:16 PM
Perhaps my method is not that good since I've made two mistakes on the same fitting. I left out a negative sign in one of my calculations for one of the planes, thus throwing off both intersections with that plane.:o

The angles on the original drawing starting from the bendline I originally solved for in my first example and working ccw are 55.169, 48.599, 48.599, 55.169, 41.897, 47.869, 47.869, and 41.897.

corrections are in bold

sharpscriber
05-27-2011, 10:47 PM
Did you correct the angle of the dangle proportional to the heat of the meat, mass of the a$$ and included the overworked rear plane of the mane on that critter? Those a$$e$$ are overworked and underpaid on some of those massive over-angled planes.
Seriously I have to make time to get to the bench. My shop is not set up and I am busy siding and roofing my home. I do find using trig hard to find out angles for this although I did well in college but forgot so much of it when I reverted back to bench layout. Give me a few days and I will certainly excite a new flame with an easy to decipher method. I just can't do it from laptop without my square, scribe, trammels, bench rule and calculator now.

sharpscriber
05-28-2011, 01:48 AM
Where does this train end? Aeviaanah that is one awesome train. I copied the pic and will make it a framed picture for my den. You're lucky to have worked on that marvelous Americana symbol.
Okay smwlocal24 I got your calculations studied and I've got to say that you've got quite an impressive intelligence with math. So if we study slope differences from one plane to another and our end result is to be 90 deg. turn at the base plane then what angle will the bend be? This is assuming to use our theoretical simple pyramid for now. Let's say that one slope is 1:2 and the other is 1:3. 3/2=1.5.

You then have to take the bend true length measurement at the corner perpendicular to the same height you measured your planes and incorporate that into your ratio.
1.5 for each of the two planes is divided into the 90 deg. so it goes into it as 60 deg.then divide by your corner measurement. And if you take the cotangent of those slope differences of 90 = (0.6666666666666667) of 90 deg. that is 60 deg. as well divided by your corner measurement. That is because you need to split the 90 deg. overall needed bend between two plane angles by two half-bends for each plane and incorporate your corner stretch measurements into it.
So even if your planes are of any slope difference the ratios will change proportionately to give you the overall bend degree calculation.
For end to end most drastic case scenarios:: A flat piece of metal will give you a 0 ratio of 90 deg which is a 0 deg. bend and full straight up duct will give you a 1 to 1 ratio which goes into the 90 deg. 90 degrees.
I measure up and over on each plane with a relevant base plane for both and find the difference between the two and process the corner factor into it. On an 8 bend model as we were given 3 views to study, very well presented by Aeviaanah, every slope to the next presented its own ratio to the next and needs to be a final 90 deg. bend to the base plane but with two bends between instead of one as on a four planed model.
Instead of one bend at 90 deg. we now have two bends at 45 deg. to ratio with or just divide the 90 deg. by three instead of two planes then add into the equation your corner measurement true lengths.. The two outside planes will intersect the next set of three planes so we will wind up with 8 in total.
Man!! No wonder why they put me on a bench all by myself in the shop. I could never explain it but it always came out right somehow. This is a very interesting topic. I vote that we should all get together and put NASA engineers out of their jobs. :laugh: (NOT!!)
I can only comprehend slope relevance to final bend needed and correlate the ratios between all three given parameters. I can't think it out as well as you do mathematically because some of my other brain parts are shorted out or lost and I forgot where I put the memory components. Maybe where I left them last. 'Been looking for a long time now.

smwlocal24
05-28-2011, 12:08 PM
First off I want to apologize to all who have tried to understand my method with what I've posted so far. I seem to have made mistake after mistake and I know how hard it is to try and learn something new from information that has a bunch of errors in it. My original post about my method is good, however my previous examples are plaqued with errors.

I used some bad values for some of my solutions which threw everything off. I redid the entire problem from start to finish and solved for all the bend angles again and this time I assure you all that I am correct. I hope no one pulled out too many hairs due to my sloppy math.

I have also uploaded a simpler problem using this method to determine the angle to bend a hip cap for a hipped roof with a 6:12 slope with 90 degree corners at the eave. I would look at this first and then the more complicated problem.

sharpscriber
05-28-2011, 12:24 PM
. I need a beer.

roofermarc
05-28-2011, 12:37 PM
First off I want to apologize to all who have tried to understand my method with what I've posted so far. I seem to have made mistake after mistake and I know how hard it is to try and learn something new from information that has a bunch of errors in it. My original post about my method is good, however my previous examples are plaqued with errors.

I used some bad values for some of my solutions which threw everything off. I redid the entire problem from start to finish and solved for all the bend angles again and this time I assure you all that I am correct. I hope no one pulled out too many hairs due to my sloppy math.

I have also uploaded a simpler problem using this method to determine the angle to bend a hip cap for a hipped roof with a 6:12 slope with 90 degree corners at the eave. I would look at this first and then the more complicated problem. My question is, why would one do all that complicated math just to find the bend angle for the cap flashing? I could draw a pitch line with a square or measure over 12" and up 6" to find the correct slope and transfer them to my folds. Of course the hip cap is broke different than the ridge and maybe impossible to just draw a line, but you can guess at it pretty good. I'll usually make a template on the job for specialized brakes of these situations. What some one needs to design is a bend calculator and mount it to our brake (hand brake) something dialed in and the blade stops at the dialed in location. I think every brake sold ought to have this mounted to it. Not the pitch angle magnetic jobber. I see an entrepreneur using my idea, or has this already been done? I have a pitch line on my shop floor that when I think I have broken to the correct angles, I hold my flashing on that template, you only have to be close and not perfect, put the flashing back in the brake and see what mark it lines up with, you know what I'm talking about. But if I had a stop dialed in mechanically it would save me a lot of time. Just saying. roofermarc out.

sharpscriber
05-28-2011, 12:53 PM
Amen brother!!! Hahaha! Can you imagine the look one the face of a shop owner who comes out to the layout bench and sees some guy (not you smwlocal24) and he's got all kinds of stuff out laying out bend calculations for stuff. Really you DO need it when you're breaking up heavy metal that can't be undone or plate. But a template of the laid out bend is what I would do from the two views of the sketch. Bends will spring back anyway. I just know where 45 deg. is and either I go more or less.
Device on a brake? We use the stop slider rod that stops the leaf. With spring back and different amounts for material thicknesses it would always be needing fine adjusting; Even same materials will change physical properties and angles will dangle.

aeviaanah
05-28-2011, 02:38 PM
First off I want to apologize to all who have tried to understand my method with what I've posted so far. I seem to have made mistake after mistake and I know how hard it is to try and learn something new from information that has a bunch of errors in it. My original post about my method is good, however my previous examples are plaqued with errors.

I used some bad values for some of my solutions which threw everything off. I redid the entire problem from start to finish and solved for all the bend angles again and this time I assure you all that I am correct. I hope no one pulled out too many hairs due to my sloppy math.

I have also uploaded a simpler problem using this method to determine the angle to bend a hip cap for a hipped roof with a 6:12 slope with 90 degree corners at the eave. I would look at this first and then the more complicated problem.
Very impressive my friend. I will start my calculus ASAP. This is really making me excited...starting off in the direction to be able to figure stuff guys in my shop cant do without a computer. Well, i and another co-worker are the only to be able to figure stuff like this with a computer anyway. I am interested in the math end of it tho. Here is an attached image, i redrafted fitting and figured angles in about 15 minutes. Cross checked with your math and you are right on the money. Your new pictures you posted are much easier to understand, I am going to print them and have a friend who has taken calculus in college try to help me out....Good job! Thank you for taking the time to figure this and correct your calculations....

aeviaanah
05-28-2011, 03:02 PM
Where does this train end? Aeviaanah that is one awesome train. I copied the pic and will make it a framed picture for my den. You're lucky to have worked on that marvelous Americana symbol.

Hey dont go framing that one...I think i have more pictures...

These pictures show before and after the jacket was installed. Ive also provided an image of what she looked like back in her prime! I'm the stud standing in the middle in one of the pictures, next to two of the parks workers. The challenging part of this project was installing it by myself, with the little help from the guys who were working on the boiler. I used ratchet straps to keep in in position...hand hammering every rivit was also quiet a blast!!

http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_2_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/2_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_3_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/3_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_4_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/4_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_5_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/5_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_jacket3May122010013_1205x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/jacket3May122010013_1205x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4190060_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4190060_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4190061_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4190061_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4260062_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4260062_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4260063_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4260063_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4260064_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4260064_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4270065_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4270065_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P4280071_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P4280071_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5030072_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5030072_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5060078_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5060078_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120104_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120104_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120105_600x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120105_600x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120109_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120109_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120111_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120111_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120113_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120113_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120115_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120115_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120118_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120118_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120119_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120119_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120120_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120120_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120121_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120121_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120122_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120122_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120123_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120123_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120124_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120124_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120136_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120136_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120142_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120142_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120144_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120144_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120147_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120147_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120152_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120152_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120153_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120153_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120155_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120155_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120156_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120156_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_P5120158_1067x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/P5120158_1067x800.jpg) http://img.photobucket.com/albums/v333/aeviaanah/Sierra/th_sierranum3final_1205x800.jpg (http://img.photobucket.com/albums/v333/aeviaanah/Sierra/sierranum3final_1205x800.jpg)

sharpscriber
05-28-2011, 03:14 PM
Well now that's a real treat. Seems like it's an adventure to be really proud of along with fun working memories for a lifetime.
Sure appreciate the pics. That is really nice of you to send to us here sweating at our keyboards salivating and chomping at the bit to drive home some of those hand rivets. I like studying details in all of these pics and even the car barn or train barn I like. Wish I was there on that one. Even to clean the railroad ballast.

smwlocal24
05-28-2011, 04:38 PM
My question is, why would one do all that complicated math just to find the bend angle for the cap flashing?

It is more useful when doing a large number of brakes where the errors would add up and possibly make a piece not work out well. For simple problems like a roof hip angle it only takes about one minute to calculate making worthwhile. However, for roof hips there is a geometric construction method for finding this angle which artisan framers would use but I believe we are limited to using this method on roofs with 90 degree eaves, while the vector method works for any geometry no matter how complex.

sharpscriber
05-28-2011, 05:04 PM
I can understand and accept that reasoning smwlocal24. Bend degrees should be calculated for a variety of applications. But isn't there any easy way to do it without a chalkboard at your layout bench looking like you're developing the formula and hypothetical explanation to the black hole or hypersonic light-wave theorem?
The givens are the end result wanted, two planes at different 0 to vector plane slopes, number of bends wanted (1 between each) and final angle at one side at least (90 deg)
Can this be measured on a layout spread with plan, profile and correlation or true length set up? Must be a connection there between the inches on squares, rules and divider spreads along with geometry radians or measurements that would work into bend degrees.
I do have a brother who has worked at NASA for over 30 years with a PHD in whatever (I forget) and he took longer than I did to layout an offset change cheek, rectangular to square fitting by using lots of math. His actually worked better than mine lol but I finished way before he did. Just the layout part.
I'll be at the bench next week and document my steps with formulas if it works. I just can't envision it without being there on a layout spread with measuring devices and going through it. You don't see those twists often.
Like that train... Wish I can build one like it lightweight and put rubber tires on it and get it street legal. Wouldn't that be awesome? Cow catcher and all.

aeviaanah
05-28-2011, 05:11 PM
Well now that's a real treat. Seems like it's an adventure to be really proud of along with fun working memories for a lifetime.
Sure appreciate the pics. That is really nice of you to send to us here sweating at our keyboards salivating and chomping at the bit to drive home some of those hand rivets. I like studying details in all of these pics and even the car barn or train barn I like. Wish I was there on that one. Even to clean the railroad ballast.
Yea this is one of the jobs i am very proud of....to be the only one from my company to work on it is also a treat. They put me on this job alone....install and everything. 4th year apprentice at the time. BTW i also placed 4th in the National competition last year. Industrial side, this was also quiet an accomplishment...2010 was a good year for me :) I won a welder, a weeks stay in vegas paid, the weeks wages, 100 a day perdium and a great experience competing. The union is a great thing for the upcoming apprentices...i have since turned out.


I can understand and accept that reasoning smwlocal24. Bend degrees should be calculated for a variety of applications. But isn't there any easy way to do it without a chalkboard at your layout bench looking like you're developing the formula and hypothetical explanation to the black hole or hypersonic light-wave theorem?
The givens are the end result wanted, two planes at different 0 to vector plane slopes, number of bends wanted (1 between each) and final angle at one side at least (90 deg)
Can this be measured on a layout spread with plan, profile and correlation or true length set up? Must be a connection there between the inches on squares, rules and divider spreads along with geometry radians or measurements that would work into bend degrees.
I do have a brother who has worked at NASA for over 30 years with a PHD in whatever (I forget) and he took longer than I did to layout an offset change cheek, rectangular to square fitting by using lots of math. His actually worked better than mine lol but I finished way before he did. Just the layout part.
I'll be at the bench next week and document my steps with formulas if it works. I just can't envision it without being there on a layout spread with measuring devices and going through it. You don't see those twists often.
Like that train... Wish I can build one like it lightweight and put rubber tires on it and get it street legal. Wouldn't that be awesome? Cow catcher and all.
Ill be waiting to see what you come up with. Mini train, im diggin it...i dont think it has to be street legal tho, lol....

smwlocal24
05-28-2011, 05:16 PM
Here is a geometric construction method I found on the web:

http://www.builderbill-diy-help.com/roofing-angles.html

I still think it would take a decent amount of math and/or drafting to be able to generate a model to measure like this.

smwlocal24
05-28-2011, 05:23 PM
BTW i also placed 4th in the National competition last year. Industrial side, this was also quiet an accomplishment...2010 was a good year for me :) I won a welder,

I placed third in the '08 competition in Vegas for architectural work. The industrial winners definitely get hooked up in those contests.

aeviaanah
05-28-2011, 05:32 PM
Here is a geometric construction method I found on the web:

http://www.builderbill-diy-help.com/roofing-angles.html

I still think it would take a decent amount of math and/or drafting to be able to generate a model to measure like this.
Interesting link, ill be sure to book mark that. Thanks.
I placed third in the '08 competition in Vegas for architectural work. The industrial winners definitely get hooked up in those contests.
Congratulations, yes the industrial winners do get hooked up. Must be companies like Miller and Lincoln using prizes as advertisements. What was your shop project? We had a blast....

sharpscriber
05-28-2011, 05:33 PM
This was a long time ago back in the 80's but I placed second in Eastern regional and first in another year. The son of one of the judges won first in my second year but I did really good. I got tongs, roper punch, books, tools and some small money. Nowadays you guys get quite a bit of 'goodies' but knowing you even went and tried is what really matters.
If I tried it today I would never compete with the newer upcoming crowd because they teach very well now. Not that I was taught poorly at all-- just newer methods.

smwlocal24
05-28-2011, 06:16 PM
What was your shop project? We had a blast....

We made an octagonal shaped copper louver. It was a cool project. I've attached a photo of the second place winner's project who scored highest on the shop portion; his project was on display at the banquet. I wish I could have gotten a picture of mine. I was very disappointed to get third after seeing his project and getting the final scores a few weeks later; I was only 10 points out of first place out of 600 possible I think.

The louver was supposed to have equally spaced vanes angled at 45 degrees. This project on display at the banquet had unevenly spaced vanes sloped about 60 degrees from the horizontal. While this contestant kept his solder neat on the bottom vane, his iron was too cold making his solder lumpy and not truly sweated into the pores of the copper. However, people do things differently all across the country so we all have our own opinions of what is best.

roofermarc
05-28-2011, 07:16 PM
We made an octagonal shaped copper louver. It was a cool project. I've attached a photo of the second place winner's project who scored highest on the shop portion; his project was on display at the banquet. I wish I could have gotten a picture of mine. I was very disappointed to get third after seeing his project and getting the final scores a few weeks later; I was only 10 points out of first place out of 600 possible I think.

The louver was supposed to have equally spaced vanes angled at 45 degrees. This project on display at the banquet had unevenly spaced vanes sloped about 60 degrees from the horizontal. While this contestant kept his solder neat on the bottom vane, his iron was too cold making his solder lumpy and not truly sweated into the pores of the copper. However, people do things differently all across the country so we all have our own opinions of what is best. I don't want to piss anybody off but that louver I dont think is right, the workmanship is great, but the blades aren't correct or the housing for that matter. He should have gotten the proven dimensions, that louver might not leak but the airflow is constricted.

sharpscriber
05-28-2011, 07:27 PM
Fail!

smwlocal24
05-29-2011, 10:46 AM
I don't want to piss anybody off but that louver I dont think is right, the workmanship is great, but the blades aren't correct or the housing for that matter. He should have gotten the proven dimensions, that louver might not leak but the airflow is constricted.

What are the proven dimensions? Most commercial louvers I've installed have had 45 degrees blades in them.

He should have constructed the louver per the shop drawings he was given which would have been at a 45 degree angle. As far as restricting airflow, I don't think it would matter much for this was intended as a gable vent with no forced air going through it. The only thing moving air across it would be the pressure gradient present from the temperature differences between inside and outside.

sharpscriber
05-29-2011, 12:25 PM
I'm sure it was just to judge an apprentice on measurements and fabrication techniques rather than having been designed as a useful item. I make my louver bottoms out of one piece anyway. No soldering at the bottom.
Back to the post at hand::: (this thread is running away with things that would be well appreciated in one of the other forums [especially those great pics!!!])
Finding the bend degrees respecting two plane alternates with a desired outcome at one of the coordinates.
You can do test pieces too and then take them back to the layout to check but what if it is expensive material or no test pieces available for testing? Or for some reason you must know the bend angle to design a component to fit that bend angle to give to another company to manufacture for you? If you are in an office without a bench available you are smart to use trig. You going to give him a broken piece of metal or are you going to give him some concrete numbers to work with? (maybe for a bracket that goes on a bend, I don't know)
HOWEVER-- The bend you are breaking up on the brake is parallel to the nose bar and leaf but on the fitting we don't think of measuring it on the perpendicular like that. We measure from the base or some relevant common point.
On this particular fitting if you put a right angle framing square on one of the obtuse angle bends SOMEWHERE that you tilt the square it will fit at all places.
Now on your layout if you put this square on certain lines at certain views of the layout you will get some angles to measure and play into formula that will give you the bend degrees.
So I am just seeing on this thread discussion that there are TWO ways to calculate this, one being mathematically and another being geometrically measured. I feel that both have equal importance and necessity for various reasons such as a phillip's screwdriver and a flat one. Some people like corn bread, others like wheat. I happen to like geometry and there is a dash of trigonometric usage there.
What dictates the bend amount (or degree)? The changes in planes and their slopes. Changes from one another has ratio. Slope changes dictate two different measurements rise over run. Then your using degrees, right? SO there are 360 in a circle. So you know you want to wind up somewhere, maybe 90 deg. around the corner. Now how do you plug all of this together so that only the bend degree is left out? Time to get to the work bench and play around with squares, trammels, rulers and a calculator. (did you know that NASA calculates with as many calculators as they can find and also hand figure their work? They will use all available mathematical processes available because of error possibilities--then decide which is correct)
Anyone seeing a connection with all of this like I do or am I just as wacky as Einstein? (mind you I'm not even a pimple on his sits)

aeviaanah
05-29-2011, 01:13 PM
I'm sure it was just to judge an apprentice on measurements and fabrication techniques rather than having been designed as a useful item. I make my louver bottoms out of one piece anyway. No soldering at the bottom.
Back to the post at hand::: (this thread is running away with things that would be well appreciated in one of the other forums [especially those great pics!!!])
Finding the bend degrees respecting two plane alternates with a desired outcome at one of the coordinates.
You can do test pieces too and then take them back to the layout to check but what if it is expensive material or no test pieces available for testing? Or for some reason you must know the bend angle to design a component to fit that bend angle to give to another company to manufacture for you? If you are in an office without a bench available you are smart to use trig. You going to give him a broken piece of metal or are you going to give him some concrete numbers to work with? (maybe for a bracket that goes on a bend, I don't know)
HOWEVER-- The bend you are breaking up on the brake is parallel to the nose bar and leaf but on the fitting we don't think of measuring it on the perpendicular like that. We measure from the base or some relevant common point.
On this particular fitting if you put a right angle framing square on one of the obtuse angle bends SOMEWHERE that you tilt the square it will fit at all places.
Now on your layout if you put this square on certain lines at certain views of the layout you will get some angles to measure and play into formula that will give you the bend degrees.
So I am just seeing on this thread discussion that there are TWO ways to calculate this, one being mathematically and another being geometrically measured. I feel that both have equal importance and necessity for various reasons such as a phillip's screwdriver and a flat one. Some people like corn bread, others like wheat. I happen to like geometry and there is a dash of trigonometric usage there.
What dictates the bend amount (or degree)? The changes in planes and their slopes. Changes from one another has ratio. Slope changes dictate two different measurements rise over run. Then your using degrees, right? SO there are 360 in a circle. So you know you want to wind up somewhere, maybe 90 deg. around the corner. Now how do you plug all of this together so that only the bend degree is left out? Time to get to the work bench and play around with squares, trammels, rulers and a calculator. (did you know that NASA calculates with as many calculators as they can find and also hand figure their work? They will use all available mathematical processes available because of error possibilities--then decide which is correct)
Anyone seeing a connection with all of this like I do or am I just as wacky as Einstein? (mind you I'm not even a pimple on his sits)
Yea i definately agree. This is the beauty of math, it proves itself over in a many many ways again and again... Id like to see you plug some numbers in heading in that direction posted above. This way i can get a better understanding of whats goin on in your noggin!

Stumbled across this....im sure youve all seen something of the sort. Not to get off topic again....

http://www.stumbleupon.com/su/7A0EBH/www.custompartnet.com/calculator/sheet-metal

sharpscriber
05-29-2011, 01:36 PM
Hey thanks dude!! That's a great stumble you found. I sure can use that. Hopefully I can find a way to plug in variables for customizing.
What I want to do with this bend degree calculating method is to make it very bare bones easy to follow. There are two or three ways to do it (without a scientific calculator) But still need to calculate simple stuff: One method scribes a radius at the corner of the plan view and some angles and works it up from there onto another elevation view; Another is to square out using the plan view bottom or base change using its angle (for non squared fittings) and measures stuff. Another uses protractor on several lines but this can get erroneous when transposing from one step to the next.
Math alone is the most precise, up to the error of the calculator being digitally controlled and operator error. But since we use our eyes to judge a line at the brake we also use it to judge our layout lines. And since the component needs to be laid out anyway I like the bench method right off the layout lines.
I have the dotted lines in my noggin but you know when you have to get to that step on the bench to see which one goes with the other. I will know when I get to follow through the steps from start to finish because I can't envision it without my layout tools.
I am so busy breaking up trim and flashing and can't get into it right now because I have to do siding and roofing and that will take me some time. But sooner or later I will either come to a stopping point or get bored with construction and go in and do it. Not that complicated but my shop is a storage bin now with EVERYTHING packed away and not organized or gone through. Not even benches set up yet, cutting by hand and using crates to work on temporarily. I'll get there; moved 4 times in the past. This time I built my own shop rather than rent.

roofermarc
05-30-2011, 09:17 PM
Hey dude, if you don't have anything nice to say then don't say it, right. What I should have said about that louver was, Thats a nice job, I just do mine different. There are no proven dimensions just common sense and imperical knowledge. Blades are always on a 45 degree. Disregard the previous post on the louver.

sharpscriber
05-30-2011, 09:22 PM
No malice intent meant at all and I know you're correct on the design part of it. For an apprentice contest it was damn good though.

cactassdupree
05-31-2011, 04:03 PM
Hey dude, if you don't have anything nice to say then don't say it, right. What I should have said about that louver was, Thats a nice job, I just do mine different. There are no proven dimensions just common sense and imperical knowledge. Blades are always on a 45 degree. Disregard the previous post on the louver.

marc Why are blades "always" at a 45 degree angle, in a louver? Don't limit yourself. you may want to adjust the angle to make the blades work out better. :) dupree

sharpscriber
05-31-2011, 04:12 PM
What I was saying about the louver was that they give a drawing sketch for the apprentices to make, not really designed for its proper air flow, blade angles, etc... Probably did that on purpose as 45 deg. was easy for that years' apprentice to follow through. They usually give them what their apprenticeship year can handle for that year. If it was for a second year apprentice it might have been just one louver and squared.

It's just to test on dimensions and fabrication. Not designed to be a proper useful thingamajig.

metalmanmania
06-28-2011, 10:40 PM
I think I have a fairly straight forward way of calculating the dihedral angle between to planes as long as you are familiar with trigonometric functions. It used as you would be calculating the pitch of the roof based of base 12. For instance 4:12, 8:12 and so on. You plug the pitch numerator( for instance 8, on a 8:12) into the equation for P. You then use it using the inverse tangent function to calculate the angle theta. From that you use that angle theta to calculate A. You now are going to use A in another inverse tangent function to calculate the angle beta. Once you have beta you simple double it and subtract the sum from 180 to get your dihedral bend angle. I derived this from scratch in 10 min so with practice this would be very quick method. I can't upload a picture of this from my iPhone so I will try to upload one from my computer.

sharpscriber
06-28-2011, 11:02 PM
Try plugging in the degrees of a slope instead of 12 based pitch ratios. What does that do when worked into an equation? 12/12 is 45 deg. and so on.

metalmanmania
06-29-2011, 12:18 AM
It does not work since you are not calculating the same angles as with which is associated with the pitch angles. You are calculating two triangles, one that is the pitch along the hip, and one for the dihedral angle, also this formula is for if both pitches are the same, if they are not a more complicated version would be required, but from my derivation you could figure out how to do that.

sharpscriber
07-08-2011, 05:24 PM
Dammit man you're good. We need a PHD college degree in sheet metal these days.

metalmanmania
07-09-2011, 01:51 PM
I went to college to be a mechancal engineer then eventually had to drop out once the kids came. The stuff I used is basic high school trig and algebra just have to apply it. I have another proof for different pitched intersection I will have to up load it when I get home

sharpscriber
07-09-2011, 03:46 PM
I'm still liking the old school way. Even if you give a fabricator some degrees to bend you still have springback, metal tension and bends will jump around because of length, lots of stuff. So I actually like the physical measurement methods. If we over broke a certain expensive project with some material designed for nuclear usage we could not fix it. It went into the dumpster.

Now if you could find a method to incorporate tension on the bending bar or die, set backs, material physics and whatever bunch of parameters to follow so that a bend could be exact as wanted then it would take way too much prep time. So physical measuring works.

I knew what dial setup would do with a set of dies in my press brake and what kind of angle it gave with a certain sheet and length. Just have to have the experience with your equipment. Dammit Jim I'm a sheet metal worker not a computer.

cactassdupree
07-09-2011, 07:40 PM
I'm still liking the old school way. Even if you give a fabricator some degrees to bend you still have springback, metal tension and bends will jump around because of length, lots of stuff. So I actually like the physical measurement methods. If we over broke a certain expensive project with some material designed for nuclear usage we could not fix it. It went into the dumpster.

Now if you could find a method to incorporate tension on the bending bar or die, set backs, material physics and whatever bunch of parameters to follow so that a bend could be exact as wanted then it would take way too much prep time. So physical measuring works.

I knew what dial setup would do with a set of dies in my press brake and what kind of angle it gave with a certain sheet and length. Just have to have the experience with your equipment. Dammit Jim I'm a sheet metal worker not a computer.

And it makes a difference in the length of the piece of work. If you make a piece 10' lg. then have to make a couple 3' lg. The degrees you brake the metal will change. :) dupree

metalmanmania
07-09-2011, 11:59 PM
I totally agree. I personally would not use this method or any other method to calculate the dihedral angle. We have charts for these for various pitches and if I was making something like a 16:12 to a 1:12 hip I would make a small model. Would only take five min and physically measure it. Then just compensate for material spring bake on our autobrake 2000. But topics like this give me something to ponder as I stand chained to the brake bending 1000s of feet of trim a day.

roofermarc
07-10-2011, 09:14 AM
I totally agree. I personally would not use this method or any other method to calculate the dihedral angle. We have charts for these for various pitches and if I was making something like a 16:12 to a 1:12 hip I would make a small model. Would only take five min and physically measure it. Then just compensate for material spring bake on our autobrake 2000. But topics like this give me something to ponder as I stand chained to the brake bending 1000s of feet of trim a day. I've made enough capping for hip roofs that I get it really close and make it work, the key is to have both a hem and a kick on it, not just a hem, like others do down here. If its a high dollar job we make samples and bring them out in the field and try/tweek them, safest way when theirs a lot on the line. Point being never rush a metal roof or a particular flashing situation.

aeviaanah
07-13-2011, 07:28 PM
It does not work since you are not calculating the same angles as with which is associated with the pitch angles. You are calculating two triangles, one that is the pitch along the hip, and one for the dihedral angle, also this formula is for if both pitches are the same, if they are not a more complicated version would be required, but from my derivation you could figure out how to do that.

Does this method work with the fitting I described in earlier posts?

metalmanmania
07-13-2011, 10:05 PM
aeviaanah.... the method that I posted before will not work for your sitiuation with the twist. It is only good for a hip or a valley on a roof when both sides are at the same pitch. However it got me thinking and I figured out a solution to your problem that is far less difficult then what it seems, no calculus required in it. The way I did it was to draw a top view of the fitting. The top view of the bend line on the twist we will call "A". From this view draw a line perpindicular to line A at the corner point at the top of the twist. The top secition of the twist we will need to draw a line perpindicular from its edge to the lower base corner, continue this line out tell it intersects the line you drew that is perpindicular to line A. On either side of this line perpandicluar to A you have to short segments, on the left measure the distance to where it intersects the lower base of the fitting, call this measurment "C". On the right side measure the distance to where this line intersects the other line you drew perpindicular to the fitting upper base, call this distance "E". Save these two measurments tell later. Now measure the length of line A, divide the height of the fitting by A and then take the inverse tangent of this number to calculate the degree THETA.
You now need to calculate B, take the SIN of Theta and multiply that by the lenght A. See the pictures if you have any questions. From there you will use the lenghts B,C,and E to calculate the Dihedral angle, The formula from this point should be pretty clear.

It isnt exactly all math since to derive a formula to calculate that for ever instance would be increadable complex but it is fairly easy once you wrap your head around it. Hope this helps you out.

metalmanmania
07-13-2011, 10:08 PM
sorry the pictures were the not straight, here is the first part of the formula, it wouldnt load the first time.