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Post subject:distance between 45 els
 Posted: Mar 09, 2004 - 03:33 AM #396
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Joined: Mar 09, 2004
Posts: 2
Location: connecticut
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| hello,i was wondering if anyone knows the formula for obtaining the length of duct between 2-45's. hope i said that right, in simple terms trying to set around a beam with 2- 45's and a 4ft joint and im tired of drawing it on the floor! |
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Post subject:RE: distance between 45 els
Posted: Mar 09, 2004 - 08:07 PM #397
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Power User
Joined: May 14, 2003
Posts: 212
Location: Sydney Australia
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Hi Clink,
The distance between the 2 ends of the duct is 2x (height of the beam you are trying to jump + clearance between the beam and the duct).
The length of the duct is the square root of (2x [height of the beam you are trying to jump + clearance between the beam and the duct squared])
This would give you a 45 degree bend at each end of the duct and a 90 degree bend under the beam. Not very efficient though. How many beams do you have to jump?
Send me a dimensioned sketch oldcramo.splinter@yahoo.com (remove the dot splinter) don't forget the duct size, and I'll give you all dimensions.
Glenn |
_________________
In Jus Voco Spurius
Home page
http://www.metalbashatorium.com
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Post subject:RE: distance between 45 els
Posted: Mar 09, 2004 - 09:08 PM #398
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| thank you for your response grue, but i dont think im saying the question correctally,let me try again with more detail. a 10x10 horizontal run of duct needs to set over 34" to avoid hitting a vertical colum,using the 2-45's with 6" throats im looking for the legth of the duct betwwen the 2 -45's. i hope that explains it better.thanks |
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Post subject:distance between
Posted: Mar 10, 2004 - 04:59 AM #399
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Power User
Joined: Feb 01, 2004
Posts: 820
Location: Mobile, Alabama
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I would search for a good applied mathimatics book it will help you greatly.
There you will find a theorum like- a2=b2=c2. If I understand right I would say your trim joint would be 32 5/8 " long. Hope this helps.
pricer pricereynolds@yahoo.com |
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Post subject:Our Link to this is
Posted: Mar 10, 2004 - 11:43 AM #401
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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Here is a link to the formula, we've had this online for some time. I am in the middle of writing a good note book on math for the sheet metal industry. And we show you how to use the calculator, I think you'll like it. It will be free for members and a slight charge for non-members
Here is the link to our one page for now
http://www.thesheetmetalshop.com/pn/ind ... amp;pid=32
It doesn't matter if it is square or round, you have to work from the centerline radius. I'll post an image of what I'm referring to more clearly. |
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Post subject:Re: distance between
Posted: Mar 10, 2004 - 06:45 PM #403
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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pricer wrote: I would search for a good applied mathimatics book it will help you greatly.
There you will find a theorum like- a2=b2=c2. If I understand right I would say your trim joint would be 32 5/8 " long. Hope this helps.
pricer pricereynolds@yahoo.com
You're correct on the length approx 32" between the elbows but the pythagorean theorum cannot be used because you still don't know the length of the base? Actually we do with a 45 deg angle, but to learn basic trig functions you will be able to solve for all angles with just your calculator in the field or in the shop. What you do know is the height of one side (34") and the angle 45 degrees
Take the calculator push 34 (your height) divided by 45 (the angle) then push the [TAN] this gives you the base length, now you can use the pythagorean theorum and only have to subtract the distance from the center line of the elbows to the end of the throat..the pythagorean theorum is from centerline (I'll link this to a drawing, because it is important that one knoows and understands this for all angles and we'll include this in our calculator paper I have almost done.
Try this with a 30 degree angle..lets say your following a wall with the duct and you want to pre make this in the shop..(without the calculator on you plasma) No Cheating now How long would you need the duct to be (without the allowance for the connection)
Bud...
More
Here is a link to a sample of our calculator papers which will include the full trig funtions and more. I'll post the one drawing in a bit and edit this page to include the link as well.
http://manual.thepiers.net/pdf_manual/Calculators.pdf |
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Post subject:distance between
Posted: Mar 10, 2004 - 09:48 PM #405
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Power User
Joined: Feb 01, 2004
Posts: 820
Location: Mobile, Alabama
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Post subject:RE: distance between 45 els
Posted: Mar 23, 2004 - 07:49 PM #452
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Joined: Mar 23, 2004
Posts: 1
Location: westport ma.
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| This is fairly easy. The formula is : the amount of offset X 1.41, then deduct the amount of straight in your 45 deg. fittings. Do this off the center lines of your ductwork not the throat or heel. P S Don,t forget to add or subtract the gain or loss in slip connections |
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Post subject:RE: distance between 45 els
Posted: Mar 23, 2004 - 10:45 PM #454
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Power User
Joined: Feb 01, 2004
Posts: 820
Location: Mobile, Alabama
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Thats the one I was trying to remember.
Thanks |
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Post subject:RE: distance between 45 els
Posted: Mar 30, 2004 - 03:02 AM #532
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| man,smoke pouring out. i laid out on a square sheet this... (no connections added) 12"pipe with a 27" offset using 2-45s with 6" throats .then measured the fill piece it was 21" i cant get 21 out of any of these formulas. c/l of the 45s were 8.25. so if i do the math like the link says 27x1.414=38.17 12x.414=4.96 x2=9.93 38.17-9.93=28.25 its off 7.25? i laid it out a little ruff. but its way off.am i wrong? |
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Post subject:RE: distance between 45 els
Posted: Mar 30, 2004 - 03:57 AM #533
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Power User
Joined: Feb 01, 2004
Posts: 820
Location: Mobile, Alabama
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It looks as though you are not working from the center line. If you are working the formula and deducting the 6" throat twice you would come up with a sum that was greater than the pipe needed. If you were to draw a side view of this problem and were to draw a line thru the center of the fitting you would find that the throat at this point was 8 1/2" (roughly).
Now if you rework the equation you will find the 21" you are coming up with. I hope I explained it well. Hope this helps.
pricer  |
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Post subject:RE: distance between 45 els
Posted: Mar 30, 2004 - 09:03 PM #537
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| the center lines were 8.25 on the above which is what i used,unless my math is wrong(that never happened before?) |
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Post subject:RE: distance between 45 els
Posted: Mar 31, 2004 - 02:14 AM #541
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| okay i understand i was using center of the 45 to the edge thats wrong but i cant firgure out the clr where that is and how you get it without drawing the fitting |
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Post subject:RE: distance between 45 els
Posted: Mar 31, 2004 - 02:46 AM #542
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Power User
Joined: Feb 01, 2004
Posts: 820
Location: Mobile, Alabama
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I believe to find that center line dimension you would need to multiply the throat length by 1.41. If you: 6” x 1.41= 8.46”.
Example: We have (2) 45 degree elbows 12” round and a length of pipe and we need to obtain an offset 27”.
27” x 1.41= 38.07”
This is the length of pipe we need but, we still need to deduct the throat dimension from this number.
To find that center line number for the throat we would multiply the throat dimension by 1.41:
6” x 1.41= 8.46”
If we now multiply this number (8.46”) by 2 (counting both throat dimensions} :
2 x 8.46”= 16.92”
Now if we were to go back to the first sum and subtract the number 16.92” (the combined throat dimensions), it should give us our answer.
38.07” – 16.92” = 21.15”
So the sum of 21.15” would be the pipe length required to do the job. As previously stated you would have to take into account the connectors.
Hope this helps, if I am wrong, please educate me too,
pricer
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Post subject:RE: distance between 45 els
Posted: Apr 01, 2004 - 09:51 PM #559
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| helps a little, i confused now on the link from bud were do you use the .414 number |
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Post subject:RE: distance between 45 els
Posted: Apr 01, 2004 - 11:50 PM #560
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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Here is what the 8 x .414 is about. the first part of the formula
L =(offset x 1.414 (provided it is a 45 degree offset) This gives you the distance center to center, then you have to minus the rest of the formula
- (centerline radius in our case 8" x .414) this give you the dim. of A to B in the drawing.
This is where that comes into the picture...Hope this helps better.
Bud |
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Post subject:RE: distance between 45 els
Posted: Apr 01, 2004 - 11:54 PM #561
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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I'm appoligize if I get behind..I'm very busy this time of year..but when I'm not we do some good stuff
Thanks for the help everyone
bud |
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Post subject:RE: distance between 45 els
Posted: Apr 02, 2004 - 02:46 PM #571
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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I'll add this to the one link so it will ba available all the time. I have much more of this type of explaination in the work I'm doing on the Calculators for Sheet Metal Workers...
Bud |
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Post subject:RE: distance between 45 els
Posted: Apr 03, 2004 - 03:02 PM #578
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| the math finally works! now to be more of a pain,this has the wheels turning and brought up two new questions,1 how do you determine the total amount of distance the 2 45s and filler piece the offset will take up end to end ? and how do you determine the offset that 2 45s back to back will be? |
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Post subject:RE: distance between 45 els
Posted: Apr 03, 2004 - 03:31 PM #579
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Webmaster
Joined: May 13, 2003
Posts: 1327
Location: Waukesha
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clinc wrote:
the math finally works! now to be more of a pain,this has the wheels turning and brought up two new questions,1 how do you determine the total amount of distance the 2 45s and filler piece the offset will take up end to end ? and how do you determine the offset that 2 45s back to back will be?
I'll put up the trig formula for this weekend sometime, but for now because it is a 45 degree angle, the offset of 12 inches would be the same as the base of that triangle and then simply add the dimension A to B instead of subtracting them The dim. A to B is the center point of the elbow to the edge..
Bud
And if your wheels are still turning, the trick question would be that now we have covered this with the understanding that this offset lies horizontal on one plane.
With the same offset what would the length of the center be if the offset had to rise 6 inches to another plane. An example would be of you were running along the wall and had to bring it up to ceiling level which is 6 inches higher.
I'll leave this alone for awhile and let you do some figuring..
Bud |
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