Author 
Message 
Chuck Deaton
Member Username: Chuckdeaton
Post Number: 31 Registered: 32002
 Posted on Thursday, June 20, 2002  7:29 pm:  
In some cases a Leica Disto can be used to determine the rafter length. If the eve is accessible and there is a target near the ridge you are in business. Also, Empire makes a protractor with a pitch calculator which used in conjuction with a steel tape or a Leica Disto will give you two angles and a side. Or a Leica Disto can be used, especially on a gable roof to measure A and B of A square plus B square equals C square which is the hypotenuse and the rafter lenght. (Rule of Pythagoras) This is explained by Paul I. Thomas in his book How to Estimate Building Losses and Construction Costs Library of Congress Catalog Card # 6013896 There is also a Paul I Thomas paperback that has a table for determing squares of roofing using the square footage and the pitch. 
Five Daily
Registered User Username: Kitchenista
Post Number: 16 Registered: 42002
 Posted on Thursday, June 20, 2002  4:51 pm:  
Todd you asked about a process to measure a roof with just the perimeter and pitch. It is called the Foot Print Method. I have attached the process. If you can't read the attachment and example let me know and I will type it out for you. Note: I think this is just as good as risking your neck on an inaccessible roof. When I use this method I document why I made the choice and I have never been questioned. Jennifer note: when you download a file from the forum it may try to save it with the unk extension, in this case just change the unk to tif, Roy

Jim Blanchard
Registered User Username: Jimb
Post Number: 4 Registered: 102001
 Posted on Sunday, May 26, 2002  10:20 pm:  
Todd  I certainly did blow it on the 4h dimension. I missed the fact that it extended beyond the rake (6a).I suspect that the areas of 5 and 6 would be close enough so that you could just double 5 and forget about 6 for estimating purposes. But as for the trapezoids (2,3,5 and 6)I am sure the formula for those areas is: the sum of the paralell sides divided by 2, times the base. You had me going though, so I had to check. Also the the Pythagorean theorum is: The SUM of the sqaures of the sides equals the square of the hypotenuse. Thus the equations should be 2a squared minus 2b squared etc. equals x squared where x is the unknown leg. Anyways you don't need Pythagorus. Subtract 1c from 3c etc. and you have the unknown leg of the triangle. All that BS said I believe your first calculations would be right on the money. And how do you think us roofers make our money? About every 5th job we don't buy any materials, so long as we can find a customer who wants a harlequin roof. I rarely have more than a couple of bundles of shingles left over on any job and then its mostly with architecturals becuase there is less waste. I always leave a full bundle with the customer as a good will gesture and because I don't want to lug any more than I have to away from the job. When I used to use the "Socks of steel" method I would have a lot left over, but now that I have my TapeDolly I can measure almost any roof (Wink). That adjuster you spoke of probably knows how cheap labor is. We cut the full rolls up just keep the men busy and create more end laps to assure call backs and claims for the adjusters.

Cecelia Sharpe
Registered User Username: Cecelia
Post Number: 8 Registered: 122000
 Posted on Sunday, May 26, 2002  5:28 pm:  
Ghost, 1984 EIGHTYfour? Good God!!! Do you have a story just a little closer to our current time frame? Perhaps in this century? Or even the last century in the LAST decade? 1984? I'm impressed with your memory. At least some of us can still remember (accurately?) that far back. 
Dale Strain
Registered User Username: Catmandale
Post Number: 16 Registered: 12002
 Posted on Sunday, May 26, 2002  4:38 pm:  
Todd, Yeah, I'll bet the roofers save all those roll ends just so they can fuss around getting them to lay flat and save a few bucks. Sure. Sometimes there are just too many pencil jockeys without a clue. (Message edited by catmandale on May 26, 2002) 
Todd Summers
Registered User Username: T4summ
Post Number: 7 Registered: 122001
 Posted on Sunday, May 26, 2002  1:58 pm:  
Jimbo, your formula would be about 5 squares too much. Typical roofer measurement procedures HA. To wit ...1b and PART of 4h are already included in 8a. And you can't simply add 1c and 3c and divide in half. Actually if you did this, your total would prabably come up short. Imagine a roofer coming up short, LOL. BTW My background is in roofing also, I'm just taking jabs at you. and you are right about the possible flat sections. Some carriers say not to add waste to rolled material, but you always have to count the number of courses and take the last course at least 18 inches up under the first course of shingles. I once had an adjuster tell me that he rounded DOWN on rolled roofing because " A big , successful roofing company always has a few butt ends laying around in their yard" !I think HE was a butt end ! LOL 
Todd Summers
Registered User Username: T4summ
Post Number: 6 Registered: 122001
 Posted on Sunday, May 26, 2002  1:40 pm:  
I've heard of a mathematical formula that exists that can determine the area of a roof knowing only 2 numbers... the total perimeter and the pitch. If this works it would be very helpful on some 2, 3 story, steep roofs. Has anyone else ever heard of this? Does it work ? Is it accurate? 
Todd Summers
Registered User Username: T4summ
Post Number: 5 Registered: 122001
 Posted on Sunday, May 26, 2002  1:34 pm:  
RJ,it's a given that this roof would require a 10% waste factor. And you are right, all of the numbers are there. It just dawned on me that you could use the pythagorean theory to figure the triangles in 2,3,5 and 6, therefore here goes... 1.1a or 1c x 1b 2.2a x 1c +(2bSquared/2asquared x 2a/2) 3.3a x 4a + (3bsquared/3asquared x 3a/2) 4.(4b x 4c)+(4h+4f X 4e)+(4h x 4g) 5.(1b x 5a) + (2bsquared/2asquared) x 2a/2 6.[8a  (2a + 1b) x 6a] +(3bsquared/3asquared x 3a/2) 7.7ax7b 8.8ax8b But the pitch rule still applies... if the pitch on 2 and 3 is steeper than the pitch on 5 and 6 this formula will overestimate. If it is viceversa, it will underestimate. Oh yeah, take the total of 1  8 and multiply bt 1.1 and round up to the bundle. 
Jim Blanchard
Registered User Username: Jimb
Post Number: 3 Registered: 102001
 Posted on Saturday, May 25, 2002  11:04 pm:  
I'll take a shot at this. The area is the sum of: 1. 1a x 1b 2. (1c+3c)/2 x 2a 3. (3c+4a)/2 x 3a 4. The sum of: 4h x 4a 4a4g x 4f 4d x 4c 5. (8a/2 + 1b)/2 x 5a 6. (8a/2 + 4h)/2 x 6a 7. 7a x 7b 8. 8a x 8b For areas that are shingles add 10% for starters, ridge cap and waste. There would appear to be a flat roof or 2 there. 10% might work there as well but it would be prudent to check dimensions against standard singleply sheet sizes etc. I have always felt that there is no such thing as too many measurements. It much easier to disregard an unneeded dimension or two than to have to return to the site to pick up a missed measurement.

Rj
Registered User Username: Rj
Post Number: 5 Registered: 122000
 Posted on Saturday, May 25, 2002  9:22 pm:  
All measurements required to accurately determine the square footage of the roof is present in Roy's drawing. However, no one has provided the correct formals to make an exact calculation. Another factor that has seemed to have been overlooked is that everyone is providing their solution for the amount of roofing without including a waste factor in their calculations. 
Ghostbuster
Registered User Username: Ghostbuster
Post Number: 280 Registered: 122000
 Posted on Saturday, May 25, 2002  6:37 pm:  
True story. Here's how the late star adjuster of Wardlaw Claim Service, Warren 'Socksof' Steele measured this roof. (I know, cause I watched him do it in 1984!) Leaving the ladder in his station wagon, he would squint up at the roof and in an authoritative voice tell Ms Insd, "Last year I was in Abilene on a roof just like yours and it was 36 1/3 squares. If your roofer has a question have him call me." Since Farmers did not do reinspections, this was considered accurately measured by a professional catastrophe adjuster entrusted with the responsible disbursement of the corporate funds. Aaahh, the good ol' days are gone forever! 
Todd Summers
Registered User Username: T4summ
Post Number: 4 Registered: 122001
 Posted on Saturday, May 25, 2002  10:51 am:  
Actually, if the pitches are not the same, you would need both measurements to calculate the areas accurately. 
Todd Summers
Registered User Username: T4summ
Post Number: 3 Registered: 122001
 Posted on Saturday, May 25, 2002  10:43 am:  
To get the accurate calculation of area, Jennifer is right. More measurements are needed to determine the areas of the triangular portions of 2 and 3. You need the base x height to divide by 2 and add to 2a x 1 c and 3a x 4a. You would also need this to subtract from (5a+6a)8a for combined area of 5 and 6 (assuming 5a is the same as 6a) However, if all pitches are the same and distance between gable and ridge is the same on 5 and 6 ...(8a/2) and you are confident enough to swag it, this formula should get you pretty close... 1.1a or 1c x 1b 2.2a x 1c 3.3a x 4a 4.(4b x 4c)+(4h+4f X 4e)+(4h x 4g) 5.(8a/2)x 5a 6.(8a/2)x 6a 7.7ax7b 8.8ax8b This will only work IF distance between gable and ridge 3c is the same in 5 and 6, and only if the pitches are the same in 2,3,5 and 6. Otherwise you would need 1 more horizontal measurement for accuracy, and it could be either 3c portion of 5 and 6 or 8a portion of 5 and 6. Todd Summers

Five Daily
Registered User Username: Kitchenista
Post Number: 15 Registered: 42002
 Posted on Thursday, May 23, 2002  7:00 pm:  
You're welcome. As for what I would do on a roof... I know it takes a few more minutes, but I always write out my calculations on my roof diagram to make sure I have everything. I don't actually calculate it, I just write out the formulas (using the actual #'s, not 5a, 1b, etc...). Sometimes I've missed a valley here or a ridge there. I'll admit, I have tried to guestimate it before in the hopes no one would reinspect me. (I've been lucky so far) That is why I write it out. On really cut up roofs I usually have to stand back up and get 1 more measurement that I missed. I think that by dividing 8a in half I would be safe on this roof. I think the math would probably come out the same even if it wasn't even. Jennifer 
Roy Cupps
Board Administrator Username: Admin
Post Number: 135 Registered: 11997
 Posted on Thursday, May 23, 2002  12:19 am:  
Thank for the great replies and thank you for sharing your knowledge. I will spend some time tonight going over them and get back to you. Jennifer, good point about 2b and 3b. Would you need to make another trip to risk or could you complete the estimate without knowing that. One problem I see from time to time is adjusters taking measurements that they don't need and of course not taking the right ones. I can remember times in the past after getting back to the hotel and sitting down to do the estimate only to realize that I didn't get the right measurement. Keeping them coming. CatAdjuster.org An Adjuster to Adjuster Community

Five Daily
Registered User Username: Kitchenista
Post Number: 14 Registered: 42002
 Posted on Wednesday, May 22, 2002  10:03 pm:  
I do not have enough info to finish my calculations. Do 2b and 3b intersect 8a exactly in the middle? If so, easy enough(1/28a), if not, then the right and left side of the 2 ridges intersecting need their own id (8c & 8d?) I see 4 trapezoids(2,3,5,6), 3 clear rectangles(1,7,8), plus section 4, which just needs to have a couple of rectangles subtracted out of it[(4d+4f)*4g & (4d+4f)*4c]. This way of computing would mean the pitch doesn't have to be a factor since each slope is broken down to a basic shape and there is no "folding in" or "flattening." Jennifer Looks like Joseph forgot to subtract 4dx4g also. 
R D Hood
Registered User Username: Old_dog
Post Number: 21 Registered: 11997
 Posted on Wednesday, May 22, 2002  5:17 pm:  
Here's My SWAG 1ax2ax2= 1cx2ax2= 4ax4b(4dx4c)(4fx4g)= 5ax2a=2b+3b= 8ax5a(5ax2a)= 8ax8b= 7a=7b= Put 30 estimators on this and you will get 30 different numbers. Also the slopes are not mentioned so if a 12/12 is diving into an 8/12 then the equasion will change to x1+y1/2xL for the various slopes. Have fun, thanks Roy 
joseph m lombardo jr
Registered User Username: Jlombardo
Post Number: 18 Registered: 12002
 Posted on Wednesday, May 22, 2002  9:38 am:  
1= 1a x 1b 2= 1c x2a +(2a x5a)/2 3= 3a x4a + (3a x6a)/2 4= 4a x4b (4f x4g) 4c x4d) 5= 5a x1b + (5a x2a)/2 6= 6a x3a + (6a x3a)/2 7= 7a x7b 8= 8a x8b 
Clayton Carr
Member Username: Clayton
Post Number: 58 Registered: 112001
 Posted on Wednesday, May 22, 2002  9:29 am:  
Roy, this is an interesting piece of math for the last coffee of the morning. I'll take the first stab at it. Please let me know what percentage I'm off (if any) plus or minus, and from which section(s). The area would be the sum of the following eight sections; (1a x 1b) (2a x 3c)  (1/2 2a x 5a) (3a x 3c)  (1/2 3a x 6a) (4h x 4g) + [(4h + 4f) x 4e] + (4b x 4c) [(5a x (1b + 2a)]  (1/2 2a x 5a) [6a x (4h  4i**) + 3a]  (1/2 3a x 6a) 7a x 7b 8a x 8b (**)4i= the length of the extension of 4h past 6a 
Roy Cupps
Board Administrator Username: Admin
Post Number: 134 Registered: 11997
 Posted on Wednesday, May 22, 2002  2:38 am:  
I'm reworking the roof diagram section of the site and thought it might be educational to see how others would measure this roof. \
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