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#1




Armstrong ceiling system calculation
Greetings
I WANT TO KNOW HOW TO CALCULATE THE QUANTITY OF MATERIAL REQUIRED AND THE THE TOTAL COSTING FOR THE GRID 2 FEET BY 2 FEET WITH 4 DIFFERENT MATERIALSMAIN RUNNER,WALL ANGLE,2FEET CROSS TEES AND 4 FEET CROSS TEES.The material cost for each of these is different and in LINEAR METRES. THANKS AMAZING 
#2




Re: Armstrong ceiling system calculation
well first you need the prices per length/ or piece to start from there you will have to get the total measurement of the ceiling and go from there.

#3




Re: Armstrong ceiling system calculation
Quote:
Edit: You may find this guide helpful: http://www.armstrong.com/common/c200...iles/15994.pdf Particularly, read carefully the part on oddwidth border pieces. If you want 2' x 2' grid, you won't use any 4' crosstees, only 2'. Don't forget hangers. I think the price per linear meter is a red herring. The cross tees are either 2' or 4' and you need to know number of pieces. The main and wall flange come in fixed lengths, 10', 12', etc. If you have them cut to special length, I doubt they absorb the scrap. Determine number of pieces and demand price per piece. I'm as prometric as anybody, but you are dealing with a footbased system. Lay it out in feet. Last edited by JohnS; 09262010 at 04:33 AM. 
#4




Re: Armstrong ceiling system calculation
hi,
Greetings I found your reply very useful and i am sure that u have understood my issue very well, Can u explain the thumb rule which states that for a 100 square metres the number of main runners required is 84 linear metres,the wall angles is 45 lm,the 4"cross tees is162 and 2"cross tees is 81 lm. Taking this to be true how would we arrive at the quantity of the 4 different materials that would be required for 100 sq metres or 1 square metre. Amazing: 
#5




Re: Armstrong ceiling system calculation
Quote:
We must establish first whether you wish to use panels that are 2 ft x 2 ft (square) or 2 ft x 4 ft (rectangular). If you use square panels, you must have main beams every 2 ft and use 2 ft cross ties. For this case, no 4 ft cross ties would be used. If you use 2 ft x 4 ft panels, there seem to be two possiblilities, depending on whether the 4 ft dimension is along the main beam or along the cross tie. If along the main beam, you will still use only 2 ft crossties, and space them every 4 ft along the main beam. If it is along the crosstie, you will use only 4 ft crossties, spaced every 2 ft along the main beam. In both cases, you need the exact room dimensions to calculate border panels (and the border beams which are incomplete pieces). With 4 foot panels, you need to determine which way the 4 ft dimension runs in the room. It will change the ratio of crosstie to mainbeam, and which dimension of crosstie is used. You need to follow the method in the installation guide and make a detailed sketch. From the sketch, determine the amounts of material needed. I strongly recommend NOT relying on that rule of thumb. It would seem to lead to errors. For example in a 100 m² room, if it is 10 m x 10 m, 40 m of wall flange is needed, but if it is 5 m x 20 m, then 50 m of wall flange is needed. 45 m is a poor approximation to both numbers (worse for the 5 x 20 room as you wouldn't have enough). The wall flange is twice the width plus twice the length. 
#6




Re: Armstrong ceiling system calculation
Greetings
HI, We use all the 4 items for any dimension room ceiling rectangular/square/any shape.The grid is always 2" by 2" as the tiles are of the same dimensions for odd shape side or corners we use cut tiles.2" BY 4" PANELS are very rarely used by us. We have found that their thumb rule is very close to the requirements,the no os pieces as per their rule for 100 square metres is as follows: Main runners = 28 nos their length is 10" each Wall angles = 16 nos their length is also 10" each 4"cross tees = 135 nos their length is 4" each 2"cross tees = 67 nos their length is 2" each. we use all the above for any shape for 100 square metres area. Can u explain it by drawing the GRID OF 1 SQUARE METRE AREA with these infomation . Thanks a lot for the prompt replies AMAZING 
#7




Re: Armstrong ceiling system calculation
Quote:
The only way I can see to use 4' crossties, and 2ft square tile is: *Run main beams 4 ft apart *Use 4' crossties between the main beams, every 2' down the beam. This results in a 2' x 4' opening *Use a 2' crosstie to divide each 2' x 4' opening into two 2' x 2' openings. From the instructions, I can't tell whether that is allowed or not. 
#8




Re: Armstrong ceiling system calculation
Greeting
Here is some litarature try making some sense and let me know how many main runners,cross tees both 2' and 4' and wall angles are required for 100 square feet or 1 square feet area to have 2 by 2 tiles PLEASE. SPECIFICATION FOR ARMSTRONG CEILINGS Providing and fixing in true horizontal level 600 mm. X 600 mm. false ceiling system manufactured by M/s. Armstrong World Industries main tee of size 24 x 32 mm., having 0.27 mm gauge at every 1200 mm. centre to centre maximum and rotary stitched cross tee of size 24 x 27 mm, having 0.27 mm gauge at every 600 mm. c/c. and sub cross tee of size 24 x 27 mm, having 0.25 mm gauge at 1200 mm c/c. and wall angle of size 19 x 19 mm., having 0.35 mm gauge fixed to the periphery of the wall. The above grid is suspended at every 1200mm c/c. in both directions using 2.0 mm. wire over the formed grid complete. amazing 
#9




Re: Armstrong ceiling system calculation
Exploring the Armstrong site, I stumbled across this material estimator. It works for rectangular or irregular rooms and you can adjust the sizes. I recommend you use it: http://www.armstrong.com/resclgam/na..._estimator.asp
It will develop a sketch of the grid layout and count up pieces for you, rounding up to the next whole length where required. For 2' x 2' tiles, it appears to be based on: Main beams on 48" centers, supported by hangers every 48" 48" Crossties, perpendicular to the main beams, every 24" along the beams 24" Subcrossties dividing each 24" x 48" opening into two 24" x 24" openings If the ceiling tiles were offered in a 600 mm x 600 mm, dimensions of 1200 mm and 600 mm could be substituted above for 48" and 24" 
#10




Re: Armstrong ceiling system calculation
Greeting
WOW THAT WAS FAST AND PRECISE GREAT HELP JOHN WOULD LIKE TO HAVE YOU IN MY FRIENDS LIST. THANKS A LOT FOR THE BURNING OF THE MIDNIGHT OIL FOR MY SAKE. AMAZING 
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