# Measure for a Roof – Down to a Science!

By Brad Caldwell – Owner of Roof, Rinse & Run – February 7, 2014 Feel free to view the original post of this article on our site: How to Measure for a Roof.

## 1. Determining Geometric Surface Area of the Roof. (â‰  Amount of Shingles Needed)

Before we show how to calculate the geometry of a roofâ€™s surface, letâ€™s reiterate that that is NOT the same as the amount of shingles needed. If a roof has an area of 40 squares (4,000 s.f.), by the time youâ€™re finished getting all the particular shingle elements, you may have 52 squares (5,200 s.f.) total. But weâ€™ll get to that later on. And are we on the same page that 1 â€œsquareâ€ (abbreviated â€œQâ€) = 100 square feet? Okay, just wanted to make sure. Letâ€™s dive in! Forgotten your high school/college geometry? Letâ€™s go over it first.

• â€œRoof Geometry Placement Testâ€

To learn how to measure for a roof, let’s first consider some basic geometry. Take a look at the lesson below. Make sense? For a simple gable, itâ€™s â€œxâ€ (eave length) times â€œyâ€ (rake length). Since thereâ€™s two sides, you multiply that by 2. For a simple hip, to find the trapezoids, you find your average measurement of â€œx1â€³ (long eave length) and â€œx3â€³ (ridge length), then multiply that by â€œyâ€ [which would be “rake length” except there is no rake, but it’s still an easy measurement to get (just measure from the eave to the ridge, with your tape laying on the shingles)]. For the triangles, multiply â€œx2â€³ (small eave length) times â€œyâ€ and divide by two. Since there are two trapezoids and two triangles, itâ€™s simple math to get the grand total. Take one last look at this â€œpre-test,â€ and if youâ€™ve got all that, weâ€™ll â€œplaceâ€ you in â€œMAT-361-R Euclidean Geometric Concepts for Roofsâ€œ!

Â â€Geometry Roof Placement Testâ€

Did you know? In Britain, â€œtrapezoidsâ€ and â€œtrapeziumsâ€ refer to the opposite things that they do in the US? Check it out â€“ weâ€™ll be using the US convention.

US British
Trapezoid A quadrilateral with at least one pair of parallel sides. A quadrilateral with no sides parallel.
Trapezium A quadrilateral with no sides parallel. A quadrilateral with at least one pair of parallel sides.
• â€œMAT-361-R Euclidean Geometric Concepts for Roofsâ€

No, you donâ€™t have to know what â€œEuclideanâ€ means, but weâ€™ll give you that as a bonus. â€œEuclideanâ€ geometry has to do with planes and anything that goes on in planes (lines, points) (Sounds like roofing, right? Calculating roofing materials is ALL about planes!). Non-Euclidean geometry would go deeper and delve into spherical and hyperbolic considerations. So for this â€œclass,â€ weâ€™ll take a look at a very complex roof and go through EACH roof plane to demonstrate how to calculate it. This is going to be the biggest step towards knowing how to measure for a roof.

This example is from a job that I bid on. After going up on the roof, I came up with the hand-sketch below. Now, it looks pretty messy, but weâ€™ll provide a computerized version in a moment â€“ that was the best I could get while standing on an 8:12 roof! (By the way, the circled numbers are estimates of square footage of shingles needed in each plane, including waste). Also, I think itâ€™s good for people to see what you actually are going to have to work with in-the-field. When you’re actually there trying to measure for a roof, you just do the best you can as far as neatness. As long as you can understand each figure, that’s the important thing. Below the messy sketch is the computer rendition, followed by its corresponding 3D version.

Â Hand-Drawn Roof Plan
Â 13 Roof Planes on This Roof
Â 3D Computer Rendering of the Roof

Okay, so did you notice that there are 13 roof planes to account for on this roof? Also, did you notice that roof plane #7 starts out about two feet (2â€²) higher than the other roof planes? (In the back, roof plane #3 also kind of started out above the others, but my Chief Architect program was having trouble figuring it out, so weâ€™ll just neglect that). So letâ€™s jump right in with roof plane #1. Weâ€™ll be looking at the math behind each roof plane.

Â Geometry of Roof Plane #1

Notice that this plane of the roof has a sort of rectangle on the left and a sort of triangle on the right. But we need to be exact, so letâ€™s divide it into three sub-regions. First, an exact rectangle on the left (28â€² x 11â€™6); then, a trapezoid with a horizontal â€œaltitudeâ€ of â€œ?â€, and two vertical â€œbasesâ€ (one is clearly 24â€², and the other one, though unmarked, is obviously the same as the 11â€™6 measurement from the nearby rake) [by the way, for trapezoids, the parallel sides which get averaged are ALWAYS called the “bases,” even if, as in this case, they are vertical; similarly, the right-angle distance perpendicular to and between those two bases is always called the “altitude,” even if it’s horizontal]; finally, thereâ€™s a triangle with a â€œbaseâ€ of â€œ??â€ and a height of 24â€². So, we just need to find what the values â€œ?â€ and â€œ??â€ are. We can agree to the following, right:

$(Equation~1)~~~ ? + ?? = 60' - 28'$

Further, because we have two right triangles (one has hypotenuse 16â€™6 and the other has hypotenuse 31â€™6), and because we know that their two furthest angles will be the same [but NOT 45Â°!]), we can equivalize as follows:

$(Equation~2)~~~ \frac{?}{16'6} = \frac{??}{31'6}$

We have two equation with two unknowns. We can substitute one INTO the other, if you remember that from your high school math class. First, we simplify Equation 2 as follows:

$(Equation~2a)~~~ ? = \frac{16'6}{31'6}*(??)$

Then, we substitute the RIGHT HAND side of Equation 2a into Equation 1 IN PLACE OF the â€œ?â€ there:

$(Equation~1a)~~~ (\frac{16'6}{31'6}*(??)) + ?? = 60' - 28'$

Simplifying this equation yields:

$(Equation~1b)~~~ (0.5238)(??) + ?? = 60' - 28'$
$(Equation~1c)~~~ (1.5238)(??) = 32'$
$(Equation~1d)~~~ ?? = 21'$

Lastly, with the number for â€œ??â€ acquired, go back to Equation 2a, and solve for â€œ?â€.

$(Equation~2aa)~~~ ? = \frac{16'6}{31'6}*(21')$
$(Equation~2ab)~~~ ? = 11'$

Couldnâ€™t we have measured for ? and ??, instead of doing the math? Yes, but sometimes thatâ€™s easier said than done. What if you donâ€™t want to go close to the edge to get a precise measurement? What if your tape persistently falls off the roof as you try to measure it? What if, standing on the ground, you canâ€™t really get an accurate starting or ending point? What are you going to secure your tape to even if you can find the exact starting and ending points? So, the point is, sometimes you need to do some math, and you might as well learn how to do it. Itâ€™s not that hard. You could also have solved for â€œ??â€ in this way, using a construction calculator and pressing the following buttons:

2 4 Feet Diag 8 Inch Pitch Run

Basically what youâ€™re doing there is telling the calculator that you have a 24â€² diagonal, an 8:12 pitch, and then by pressing the last button â€œRun,â€ youâ€™re asking for the length of the run, which it returns to you as 19â€²-11 5/8â€³. By geometry, THAT â€œRunâ€ will happen to be equal to our â€œ??â€ figure (to see that visually, skip down to the large hand-sketch of the pyramid). Now, you say, â€œWait a minute! Didnâ€™t we find our unknown, â€˜??â€™ to be 21â€²?â€ Youâ€™re right â€“ we did. The problem is that my 60â€² measurement was too conservative (itâ€™s hard to measure a long length like that with one person) and probably should have been 59â€² or maybe just a wee less than that. Then, when you reiterate the formulas above, you get something like 20â€™4, which is only about 4â€³ off from our construction-calculator (and probably more accurate) measurement. When youâ€™re in the field, you do your best, but some calculations are going to have to be â€œrounded up.â€ Anyway, I just used 21â€² in the area calculations below.

Â Two Ways to Find â€œEave Runâ€ Below a Hip â€“ Using Slant Height & Roof Pitch; Or, Using Hip Length & Hip Pitch (â€œTriangle-2â€³ Type)

Now that weâ€™ve got â€œ?â€ and â€œ??,â€ we can plug them back into the equations we came up with for each of the sub-parts of this roof plane, to get 322 s.f. for the rectangle, 195 s.f. for the trapezoid, and 252 s.f. for the triangle on the far right. If you want an â€œexercise,â€ crunch the numbers for yourself, and see if you get the same thing. So, adding all that up gives us a grand total of 769 s.f. for roof plane #1. Most of the other planes will be much easier!

$A_{1} = 769 s.f.$

Now, one last point before moving onâ€¦ Were you tempted to think that â€œ??â€ would have been 24â€², the same as the other leg, since itâ€™s a right triangle and the other angles looked to be 45Â°? You may have been further inclined to think thusly in considering that the pitch on both sides of the hip on the right side of the triangle was the same (8:12). But the problem is that youâ€™re looking at the birdâ€™s-eye-view diagram like itâ€™s 2D. In reality, the roof is a 3D structure, and what is 45Â° on the birdâ€™s-eye view actually happens to be 48.8Â° on that roof plane. Furthermore, we know that the true line of the hip goes up at a little less than 6:12 for an 8:12-pitched roof. However, if we were to rip off that triangular plane from off the rest of the roof, and stand it perfectly upright, then the hip line would have a pitch of about 15:12 being stood upright (which it never is â€“ just Iâ€™m trying to show you what it is RELATIVE to that TRIANGLE). So, for an 8:12-pitched roof, the hips will be at (a little less than) 6:12; but, detached, stood upright, and relative to itâ€™s triangle, the hipâ€™s â€œpitchâ€ is about 15:12. In other words, there are at least three triangles to keep straight in your mind! As you measure for a roof, keep these triangles separate.

Â Three Geometrically Separate Triangles to Keep Straight in Roofing

So letâ€™s look into why our previous unknown â€œ??â€ could not be equal to 24â€² (the â€œslant heightâ€). Consider first that the two angles of the triangle, while looking the same, are quite different. Hereâ€™s an extreme example. I was going to show you a 240:12 pyramid roof, but Chief Architect apparantly can only specify a maximum pitch of 68:12. Also, I was going to show you a 50â€² x 50â€² building, but itâ€™s difficult to draw the exact measurements (at least for me). Still, you can see the gist of what I was getting at â€“ in the 2D â€œbirdâ€™s-eyeâ€ view, âˆ a and âˆ b appear to be identical (i.e., 45Â° each); however, in the 3D view, itâ€™s obvious that âˆ a is much smaller than âˆ b, and as you approach a perfectly-vertical, infinitely-high spire, âˆ a would approach 0Â°, and âˆ b would approach 90Â°, while STILL appearing to both be equal (45Â°) in the 2D â€œbirdâ€™s eyeâ€ diagram! Thatâ€™s why roofers have to be careful to remember what PLANE they are in, and not get the 2D and 3D aspects mixed up!

Â 2D Deceptive Angles to Watch Out For: A Ridiculously Steep Roof (68:12)

Anyway, Chief Architect is a very advanced program, and if you learn enough about it, Iâ€™m sure you can get past the snafus Iâ€™m running into. You can check it out at www.chiefarchitect.com. But since I wasnâ€™t able to use Chief to create what I wanted, hereâ€™s the hand sketch. It better illustrates the idea of just how different an angle can look in 2D versus 3D.

Â 2D Deceptive Angles to Watch Out For: An Insanely Steep Roof (240:12)

So actually the sketch was 100â€² x 100â€² (not 50â€² x 50â€²), but you get the idea. Itâ€™s exactly 1,000â€² tall, providing exactly a 240:12 pitch (donâ€™t try to walk on that!). Notice that while âˆ a and âˆ b appear to be the same (45Â°) in the Birdâ€™s-Eye View, they are QUITE different in reality (as seen in the Elevation View)! In the proper plane, âˆ a = 2.86Â° and âˆ b = 87.14Â°! Notice also that where the hip is, there are TWO measurements, depending on what youâ€™re talking about. If your talking about the true HIP measurement, itâ€™s 1,002â€™6â€³. If youâ€™re talking about the SLANT HEIGHT (which lies underneath the HIP at least on my (2D) Elevation View), itâ€™s 1,001â€™3â€³. Confused? Take a look at the next photo to get it straightened out.

Â More Than One Triangle in a Pyramid!

Note that triangles 1 and 4 have the same measurements (unless you have a â€œswitch pitch roof,â€ where the front-back pitch is one thing, and the left-right pitch is (usually) steeper). Notice that thereâ€™s a difference between SLANT HEIGHT, HEIGHT, and HIP. Those are the three main things that can get easily confused on a roof. Note that for the roof planes in our â€œinsanely steepâ€ roof, the pitch is 240/12. However, if you were to take the right half of â€œtriangle 2â€³ out of our â€œinsaneâ€ roof, and stand it upright (it already is almost vertical), the â€œpitchâ€ of the â€œhipâ€ (RELATIVE TO THAT TRIANGLE, ONLY) would be 240 1/ 8 :12 (a tiny change). Remember when we ripped out the plane from the roof I bid on (8:12) and stood it up? We found that the hipâ€™s RELATIVE pitch, stood-upright, was about 15:12 (a HUGE change from 8:12). The point is, unless youâ€™re dealing with insanely steep roofs, the hipâ€™s â€œtwoâ€ pitch values (True-relative-to-gravity and theoretical-relative-to-stood-up-roof-triangle) will be wildly different. Just make sure not to confuse the two. Further, if you were interested in finding the true pitch of the HIP (i.e., if you could â€œwalkâ€ right along itâ€™s edge), it would actually be 170:12 (SIGNIFICANTLY LESS â€“ but still not safe to walk on!). How did I get that? Well, in the base of the pyramid, itâ€™s 50â€² both ways to the center point. So, just use the Pythagorean Theorem to find that base E = 70â€² 8 1/2â€³. Then, the height for that triangle is 1000â€². Revisit Pythagoras and you get a pitch of roughly 170:12. Thatâ€™s why itâ€™s easier, by sheer geometry, to climb a hip or a valley rather than a plain roof plane.

Always put safety first, when you measure for a roof. Whether it’s loose granules, a steep-pitch roof, “peer-pressure” from the home owner to look “manly,” or a wide-open steep roof with few or on valleys; you want to be safe. As the roofing saying goes, no roof is important enough to get hurt over. Safety, when you measure for a roof, is more important than when you’re actually working. When you’re actually working, you’re free to easily anchor yourself to the ridge, you have other people close by to help you if you’re in a bind, etc. But when you measure for a roof on your own, you’ve really got to be careful. In fact, if it just doesn’t seem safe, why not order an aerial plan from EagleViewTech? We’ll cover them later in the post.

So we said that hips and valleys are safer to walk than “wide-open planes.” But further, why is it easier to climb a valley of the same pitch than a hip? Often they are the same pitch, so why do valleys feel so much safer? Itâ€™s because for a hip, youâ€™re actually standing on roof planes that are trying to throw you off away from the hip; whereas with a valley, youâ€™re standing on roof planes that are trying to throw you into the valley. So essentially, with a valley, you are (1) constrained to the easier, lower pitch, and (2) you sort of have â€œhand-railsâ€ (the two roof planes ascending out of the valley) to protect you further. You practically canâ€™t fall when climbing a valley (except backwards [i.e., if the valley’s pitch is too steep for you]). For a hip, if the roof is 10:12, youâ€™re really just climbing a 10:12 with the added comfort of something to sort of hang on to (the hip) â€“ because each step you take places your full body weight on a 10:12 roof plane that is trying to throw you off the hip. Unless youâ€™re sitting on it and scootching up, then itâ€™s a 10:12 with a somewhat-handrail, not a 7:12 (which is, geometrically, the pitch of that imaginary line, the â€œhipâ€ [remember, it’s really a LINE, not a PLANE]). By the way, hereâ€™s a chart of ACTUAL pitches of hip/valley LINES for common roof pitches, as considered in two triangular planes. Additionally, we look at converting roof pitch to degrees, and their incremental values.

Â Figuring Hip Pitches Relative to Roof Pitch; Converting Roof Pitch to Degrees

Notice that each time you add two more inches to the â€œrise,â€ the successive amount of degrees added gets smaller and smaller. However, as you approach the fall limit, tiny changes in angle can have a large impact on your footing. Now would be a good place for a break if you need one. Before we continue, let me highly recommend the Construction Master Pro calculator. Itâ€™s an invaluable tool for construction, as you can plug feet-and-inch values into it, and it automatically corrects them (in the background) to some standard aspect when multiplying, etc., so that you can stay in the â€œfeet-and-inchesâ€ mindset and keep perfect accuracy. It also has buttons for rise, run, diagonal, pitch, etc. Itâ€™s a must-have for roofers. Itâ€™s extremely user-friendly and Iâ€™ve never been frustrated with it. Somebody thought before making that calculator â€“ you canâ€™t posit such an accolade on much of anything nowadays. And if you have a smart phone, I would get the Construction Master Pro app. Itâ€™s even better than the handheld calculator â€“ and youâ€™ll have it with you wherever you are! An absolutely essential tool if you’ve got to measure for a roof!

Â An Excellent Construction Calculator

Okay, back to calculating our roof planes.

Â Geometry of Roof Plane #2

Well, this roof plane is easy, right? Itâ€™s a trapezoid. Just plug in the two bases and the altitude (height).

$(Equation~3)~~~ A_{trapezoid}=\frac{b_{1}+b_{2}}{2}*a$
$(Equation~3a)~~~ A_{2} = \frac{(28' + 7'6)}{2}*11'6 = 204 s.f.$

Now, in real life, this roof plane was slightly different at the bottom, but weâ€™ll neglect that due to the difficulty with Chief. Sometimes on the field you will have to kind of â€œestimateâ€ a small area thatâ€™s not very conducive to a geometric shape. In real life, roof plane #2 went back under roof plane #3 a little, but again, the number weâ€™re getting here is really close.

Â Geometry of Roof Planes #3 and #4

So for roof plane #3, itâ€™s just another trapezoid (by the way, this roof plane alone was a 3:12 instead of 8:12):

$(Equation~3)~~~ A_{trapezoid} = \frac{b_{1} + b_{2}}{2}*a$
$(Equation~3b)~~~ A_{3} = \frac{51'+34'}{2}*22'6=956 s.f.~~~(Area~of~Roof~Plane~No~3)$

However, for roof plane #4, itâ€™s what? Thatâ€™s right â€“ a trapezium! I donâ€™t know any formula for that, so weâ€™ll break it down into a trapezoid plus a tiny triangle (note the tiny dotted blue line separating the tip triangle from the newly formed trapezoid below). Note that we have three unknowns here â€“ ?, ??, and ??? â€“ but Iâ€™m fairly confident we can find them all. Iâ€™m going to borrow some math from roof plane #3 to help us get variable â€œ??.â€ In the picture, it looks like roof plane #4 goes past plane #3, but not in real life (check on my messy hand-drawn version to verify). So, we can simply do this to solve for â€œ??â€:

2 2 Feet 6 Inch Diag 3 Inch Pitch Run

That spits out ?? = 21â€²-9 15/16â€³.

$?? = 21' 9~15/16"$

Then we can write equation 4 to solve for â€œ?â€:

$(Equation~4)~~~ ? + ?? = 24'$
$(Equation~4a)~~~ ? = 24' - 21'9~15/16"$
$(Equation~4b)~~~ ? = 2' 2~1/16"$

Finally, since the tiny triangle is a right triangle, we may apply Pythagorus:

$(Equation~5:~Pythagorean~Theorem)~~~ a^2 + b^2 = c^2 ~~~(where~c~is~the~hypotenuse)$
$(Equation~5a)~~~ {(2' 2~1/16")}^{2} + {(???)}^{2} = {(2'6)}^{2}$

A bit of math on the construction calculator:

2 Feet 6 Inch Conv

x2%

âˆ’ 2 Feet 2 Inch 1 / 1 6 Conv

x2%

=

1.532959 SQUARE FEET
Conv

âˆšxClear
1â€²-2 7/8â€³

So we may say the following:

$(Equation~5b)~~~ ??? = 1' 2~7/8"$
$? = 2' 2~1/16" ~~~?? = 21' 9~15/16" ~~~??? = 1' 2~7/8"$

Plugging these numbers into simple equations for triangles and trapezoids yields a total area for roof plane #4 of 146 s.f.

$A_{4} = 146 s.f.$

If we had approximated the area as simply one big triangle with two legs (24â€²) and (12â€²), we would have come up with 144 s.f., which wouldnâ€™t have been very off. Iâ€™ve provided this math, though, to demonstrate how to divide up a trapezium, in case you run into a large trapezium-roof-plane that you need to calculate exactly. Moving right alongâ€¦

Â Geometry of Roof Plane #5: A What-shall-me-call-it Nonagon

Notice that Iâ€™ve dissected roof plane #5 into eight clear geometric shapes (rectangles, trapezoids, and triangles) labeled capital A through H. Also, instead of solving for unknown question marks this time, Iâ€™ve labeled our seven unknowns as lower case a through g. First point. aâ‰ b. I was tricked into thinking, for a moment, that a=b, even after all the examples on steep pyramid roofs to prove otherwise! By the way, unless you have a FLAT ROOF (or a SWITCH-PITCH ROOF), when youâ€™re looking at a birdâ€™s-eye (plan) view of a roof, leg a of a right triangle will NEVER equal leg b! Okay, back to our math â€“ we can do this:

1 8 Feet 8 Inch Diag 1 4 . 5 Inch Pitch Run
$11'-10~13/16" ~~~( = b)$
Rise
$14'-4~9/16" ~~~( = a)$

Now youâ€™re probably wondering where I came up with a 14.5:12 pitch, right? Well, remember that triangle A is a â€œtype 2â€³ triangle; that is to say, you can go back to the table for hip/valley pitches, and take the proper hip pitch for an 8:12 roof. How did I know that triangle A was â€œtype 2â€³? Because thatâ€™s the triangle that forms the surface of a pyramid â€“ the other two (or three) triangles had to do with the INTERIOR of the pyramid. But our triangle â€œAâ€ is on the surface of the pyramidal shape, so we have to use the values for â€œtype 2â€³ triangle. In fact, ANY TIME you only know the HIP length (and donâ€™t know the SLANT HEIGHT) and you need to find the length of â€œrunâ€ that goes along the eave line (or along the pitch-transition-line, as here), you need to know these â€œtriangle 2â€³ pitches (so Iâ€™d recommend writing them down â€“ Iâ€™ll even give you the exact values for better accuracy).

*The Hip Length and Hip Pitches Below DO NOT APPLY in the Case of a Switch-Pitch Roof!
Roof Pitch Slant Height (for a RUN value of 12â€²) *Hip Length (for a RUN value of 12â€²) *Hip Pitch (relative to triangle 2) (relative to eave/transition-run) (Inches : Inches) *Hip Pitch (relative to triangle 3) (what it would feel like to walk on) (Inches : Inches)
16:12 20â€² 0â€³ 23â€² 3 7/8â€³ 20 : 12 11 5/16 : 12
14:12 18â€² 5 1/4â€³ 22â€² 0â€³ 18 7/16 : 12 9 7/8 : 12
12:12 16â€² 11 5/8â€³ 20â€² 9 3/8â€³ 17 : 12 8 1/2 : 12
11:12 16â€² 3 3/8â€³ 20â€² 2 11/16â€³ 16 5/16 : 12 7 3/4 : 12
10:12 15â€² 7 7/16â€³ 19â€² 3 3/8â€³ 15 5/8 : 12 7 1/16 : 12
9:12 15â€² 0â€³ 19â€² 2 1/2â€³ 15 : 12 6 3/8 : 12
8:12 14â€² 5 1/16â€³ 18â€² 9 1/8â€³ 14 7/16 : 12 5 11/16 : 12
7:12 13â€² 10 11/16â€³ 18â€² 4 1/4â€³ 13 7/8 : 12 4 15/16 : 12
6:12 13â€² 5â€³ 18â€² 0â€³ 13 7/16 : 12 4 1/4 : 12
5:12 13â€² 0â€³ 17â€² 8 5/16â€³ 13 : 12 3 9/16 : 12
4:12 12â€² 7 13/16â€³ 17â€² 5 1/4â€³ 12 5/8 : 12 2 13/16 : 12
3:12 12â€² 4 7/16â€³ 17â€² 2 13/16â€³ 12 3/8 : 12 2 1/8 : 12
2:12 12â€² 2â€³ 17â€² 1 1/16â€³ 12 3/16 : 12 1 7/16 : 12
1:12 12â€² 0 1/2â€³ 17â€² 0â€³ 12 1/16 : 12 0 11/16 : 12

Once you understand that, itâ€™s easy â€“ â€œaâ€ is simply the â€œrise,â€ and â€œbâ€ is simply the â€œrun.â€ So, with those values found, the area for triangle â€œAâ€ can be found:

$(Equation~6) ~~~A_{triangle} = \frac{1}{2}*b*h ~~~(where~b = "base" = our~"b";~and~h = "height" = our~"a")$
$(Equation~6a) ~A_{A} = \frac{1}{2}*(11' 10~13/16)*(14' 4~9/16) = 85.57 s.f.$

Now that we know â€œa,â€ we can find â€œcâ€:

$c = 16'6" - a = 16'6" - 14'4~9/16" = 2' 1~7/16"$

Then, g = b â€“ (2â€™6â€³ converted-to-run-value). So, plug 2â€™6â€³ through the exact same process as 18â€™8â€³ to arrive at a â€œrunâ€ value of 1â€² 7 1/8â€³. Then:

$g = 11'10~13/16" - 1' 7~1/8" = 10' 3~11/16"$

Then, find area in trapezoid B:

$(Equation~3)~~~ A_{trapezoid} = \frac{b_{1} + b_{2}}{2}*a ~~~(where~a="altitude"~or~"height")$
$(Equation~3c)~~~ A_{B} = \frac{(11'10~13/16 + 10'3~11/16)}{2}*(2'1~7/16") = 23.54 s.f.$

Next, rectangle C is easy:

$(Equation~7)~~~ A_{rectangle} = b*h ~~~(where~b = "base,"~and~h = "height")$
$(Equation~7a)~~~ A_{C} = (14')*(16'6") = 231 s.f.$

Unknown â€œdâ€ is simply the hip measurement (2â€™8â€³) converted to a â€œrunâ€ value:

2 Feet 8 Inch Diag 1 4 . 5 Inch Pitch Run
$1' 8~3/8" ~~~( = d)$

Trapezoid D is easy now:

$(Equation~3)~~~ A_{trapezoid} = \frac{(b_{1} + b_{2})}{2}*a$
$(Equation~3d)~~~ A_{D} = \frac{(14'6 + 16'6)}{2} * (1'8~3/8") = 26.3 s.f.$

Rectangle E is easy:

$(Equation~7b)~~~ A_{E} = (6'6)(14'6") = 94.25 s.f.$

We already calculated the â€œrunâ€ value for a 2â€™6 hip, so:

$e = 1' 7~1/8"$

Then obtain trapezoid F:

$(Equation~3)~~~ A_{trapezoid} = \frac{(b_{1} + b_{2})}{2}*a$
$(Equation~3e)~~~ A_{F} = \frac{(14'6 + 13')}{2}*(1'7~1/8") = 21.9 s.f.$

Grab rectangle G:

$(Equation~7c)~~~ A_{G} = (6')(13') = 78 s.f.$

Last but not least, triangle H. First, perform the â€œfunnyâ€ operation on hip â€œ16â€™6â€³ to convert it to the relative â€œrunâ€ value:

1 6 Feet 6 Inch Diag 1 4 . 5 Inch Pitch Run
$10' 6~1/4" ~~~;( = f)$

Plug it in for triangle H:

$(Equation~6) ~~~A_{triangle} = \frac{1}{2}*b*h ~~~(where b = "base" = our "f"; and h = "height" = our 13')$
$(Equation~6b) ~~~A_{H} = \frac{1}{2}*(10' 6 1/4")*(13') = 136.76 s.f.$

Combine all the areas for each geometric subdivision of roof plane #5:

$(Equation~8)~~~ A_{5} = A_{A} + A_{B} + A_{C} + A_{D} + A_{E} + A_{F} + A_{G} + A_{H}$
$(Equation~8a)~~~ A_{5} = 85.57 + 23.54 + 231 + 26.3 + 94.25 + 21.9 + 78 + 136.76 = 697 s.f.$
Â Geometry of Roof Plane #6

Similar to roof plane #1, this time it should go lots quicker! Notice that the left rake (13â€™6) is smaller than the right one (15â€™3). Rectangles A and D are too easy:

$A_{A} = (24')(13'6) = 324 s.f.$
$A_{D} = (18')(15'3) = 275 s.f.$

Now, to find altitudes â€œaâ€ and â€œbâ€ of trapezoids B and C, we cannot use the slant height (because the hips donâ€™t go all the way down â€“ it would provide a value almost twice as large as what we need); but, we can use the two hip lengths with their pitches (â€œtriangle 2 typeâ€). To find a and b:

1 8 Feet 8 Inch Diag 1 4 Inch 7 / 1 6 Pitch Run
$11' 11~3/16" ~~~( = a)$
1 6 Feet Diag 1 4 Inch 7 / 1 6 Pitch Run
$10' 2~3/4" ~~~( = b)$

By the way, did you notice I used the more accurate (14 7/16 : 12) pitch this time, instead of just (14.5 : 12) as I had done earlier? Our findings for â€œaâ€ and â€œbâ€ would have been 3/8â€³ different (in error), otherwise. Now, letâ€™s find the area of trapezoids B and C, and the total area of roof plane #6:

$A_{B} = \frac{(13'6 + 27'8)}{2}* (11' 11~3/16") = 246 s.f.$
$A_{C} = \frac{(15'3 + 27'8)}{2}*(10' 2~3/4") = 220 s.f.$
$A_{6} = A_{A} + A_{B} + A_{C} + A_{D} = 324 + 246 + 220 + 275 = 1,065 s.f.$
Â Geometry of Roof Plane #7: A Katy-Wompus Hexagon!

Donâ€™t you love large katy-wompus roof planes like this one? Not infrequently, as you measure for a roof, you’ll run into them. Theyâ€™re critical to get right, yet how do you do it?? Well, weâ€™re going to break this hexagon down piece-by-piece. So for this roof plane, I had to do it by hand. Chief Architect, my drafting program, was having so much trouble understanding that this roof plane started 2â€² higher than the other ones that it resulted in a shape that was geometrically different than reality. Notice on this weird roof plane that there are 3 sets of parallel lines, but none of the lengths are the same. Iâ€™m going to split it up into six geometric shapes (A through F). Then, we have ten unknowns to solve for (a through j). Letâ€™s find â€œaâ€ and â€œbâ€ first:

2 4 Feet 6 Inch Diag 1 4 Inch 7 / 1 6 Pitch Run
$15' 7~15/16" ~~~( = a)$
Rise
$18' 10~1/8" ~~~( = b)$

Then we can find c:

$c = TOTAL SLANT HEIGHT - b = 25' 6 - 18' 10~1/8" = 6' 7~7/8"$

Finding â€œdâ€ is a little tricky â€“ â€œdâ€ will be equal to â€œaâ€ minus the â€œRUNâ€ value of the 9â€²-hip (using â€œtriangle 2â€³ value). First find the funny â€œRUNâ€ value:

9 Feet Diag 1 4 Inch 7 / 1 6 Pitch Run
$5' 9~1/16" ~~~( = "RUN")$

Then find d:

$d = 15' 7~15/16" - 5' 9~1/16" = 9' 10~7/8"$

Subtract d from the total eave-length of this roof plane to get e:

$e = 23' 6" - 9' 10~7/8" = 13' 7~1/8"$

Letâ€™s skip â€œfâ€ for a moment and go for g, h, & i. By geometry, i=g, so we can just solve for g to get both of those. Weâ€™ll use the hip length and apply the â€œfunnyâ€ (triangle 2) pitch:

1 6 Feet Diag 1 4 Inch 7 / 1 6 Pitch Rise
$12' 3~5/8" ~~~( = g = i)$
Run
$10' 2~3/4" ~~~( = h)$

Having those numbers,â€fâ€ is simply g + i, or even more simply, 2(g), since g and i are equal:

$f= 2(12' 3~5/8") = 24'7~1/8"$

And â€œjâ€ is just the ridge length minus â€œeâ€:

$j = 14' - e = 14' - 13' 7~1/8" = 0' 4 7/8"$

Letâ€™s solve for our areas, and then for the total area of this hexagonal roof plane:

$A_{A} = \frac{1}{2}*(15' 7~15/16")(18' 10~1/8") = 148 s.f.$
$A_{B} = \frac{(15' 7~15/16" + 9' 10~7/8)}{2}*(6' 7~7/8") = 85 s.f.$
$A_{C} = b*h = (13' 7~1/8)(25' 6") = 347 s.f.$
$A_{D} = \frac{(25' 6" + 24' 7~1/8")}{2} * (0' 4~7/8") = 10 s.f.$
$A_{E} = A_{F} = \frac{1}{2}*(10' 2~3/4")(12' 3~5/8") = 63 s.f.$
$A_{7} = A_{A} + A_{B} + A_{C} + A_{D} + A_{E} + A_{F} = 148 + 85 + 347 + 10 + 63 + 63 = 716 s.f.$
Â Geometry of Roof Plane #8: A Hip-to-Valley Run with a Gable

On this roof plane, c = 2â€™8â€³ (look at the hip measurement above). Also, by geometry, b = c = 2â€™8â€³. We can say that here, because even if you made the roof slope something insanely steep, like 240:12, b would be huge, and c would be huge, but theyâ€™d both still be equal. Itâ€™s not the same situation as we had earlier. Finding a is also easy:

$a = 34' - 24' 6" = 9' 6"$

Thereâ€™s one variable I forgot â€“ weâ€™ll call it â€œdâ€ and it would be a horizontal line that would strike through the letter capital C. Itâ€™s the â€œslant heightâ€ of triangle C, and to find it, we imagine that line b in our drawing is a hip, and apply that â€œfunnyâ€ operation again:

2 Feet 8 Inch Diag 1 4 Inch 7 / 1 6 Pitch Rise
$2' 0~5/8" ~~~( = d)$

The reason we canâ€™t say A = 1/2(b)(c) is because the angle between them is not 90Â°. It is 90Â° in 2D, but as you lift that point higher and higher, the angle becomes smaller and smaller. Iâ€™m not sure what the angle is at our 8:12 pitch, but the point is, we canâ€™t use those legs. We have to use the 4â€² eave as the base, and then use the SLANT HEIGHT as the height. Then, find our areas:

$A_{A} = (34')(2' 8") = 91 s.f.$
$A_{B} = \frac{1}{2}*(8')*(9' 6") = 38 s.f.$
$A_{C} = \frac{1}{2}*(4')(2' 0~5/8") = 4 s.f.$
$A_{8} = A_{A} + A_{B} + A_{C} = 91 + 38 + 4 = 133 s.f.$
Geometry of Roof Plane #9: A Parallelogram!

This oneâ€™s easy â€“ itâ€™s a parallelogram! Just base times height to find the area. The only catch is, make sure you use the correct â€œheight.â€ You want the distance between one set of parallel lines (either set), but it has to be measured PERPENDICULAR to those lines. So, if you try to use 6â€™6 as the â€œheight,â€ youâ€™ll get a wrong answer. The height we want is 4â€™6:

$A_{9} = (34')*(4' 6") = 153 s.f.$
Â Geometry of Roof Planes #10 & 11: Trapezoid and Parallelogram
$A_{10} = \frac{(31' 6" + 34')}{2}* (2' 6") = 82 s.f.$
$A_{11} = (31' 6")*(5') = 158 s.f.$
Â Geometry of Roof Planes #12 & 13

Hopefully by this point, youâ€™re getting the feel for things. Iâ€™m dividing roof plane #12 into two triangles, whilst dividing roof plane #13 into a rectangle and a triangle. Weâ€™re going to solve for a, b, and c first, then find areas. I trust you will be able to follow along without much commentary this time. Let me just mention, though, the 8:12 pitch is used now, not the 14 7/16 : 12 (think about the pyramid and itâ€™s angles, and you should understand what Iâ€™m doing). The roof pitch has always been 8:12 (except for roof plane #3, which was a 3:12), but the situation is different here. I hadnâ€™t needed to solve for the SLANT HEIGHT in the other examples much, but here I do. To find a:

2 Feet 6 Inch Run 8 Inch Pitch Diag
$a = 3' 0~1/16"$
$b = RIDGE - EAVE = 18' - 7' 6" = 10' 6"$

To find c:

1 0 Feet 6 Inch Run 8 Inch Pitch Diag
c = 12â€² 7~7/16â€³
$A_{12} = \frac{1}{2}*(8')*(9' 6") + \frac{1}{2}*(2' 6")*(3' 0~1/16") = 38 + 8 = 46 s.f.$
$A_{13} = (7' 6")*(15' 3") + \frac{1}{2}*(10' 6")*(12' 7~7/16") = 114 + 133 = 247 s.f.$

Finally, we find the total area for the whole roof!!

$A_{Total Roof} = A_{1} + A_{2} + A_{3} + A_{4} + A_{5} + A_{6} + A_{7} + A_{8} + A_{9} + A_{10} + A_{11} + A_{12} + A_{13}$
$A_{Total Roof} = 769 + 204 + 956 + 146 + 697 + 1,065 + 716 + 133 + 153 + 82 + 158 + 46 + 247$
$A_{Total Roof} = 5,372 s.f. = 53~2/3~(say)~54~Q~(squares)~ (of~Architect~Shingles)$

Okay, if you really followed me along for all of that, weâ€™ll say that you successfully passed â€œMAT-361-R Euclidean Geometric Concepts for Roofsâ€!

By the way, you can order roof plans almost overnight from â€œEagle View Technologiesâ€ for a very reasonable rate (I got an aerial view of this house for $55). Guess what they came up with for total geometric roof area? 5,376 s.f.! Can you believe that? Only 4 s.f. of difference from my calculations, just by using aerial photos and some technology! Iâ€™d highly recommend them! If youâ€™ve got a dangerous roof to measure,$55 is not a lot of money to keep your feet on the ground! You’re able to measure for a roof and be safe at the same time. Youâ€™d probably spend \$55 of your own labor just getting the measurements, anyway! Also, even though Eagle View Technologies gives a loud disclaimer about how, if they provide you with a 2D extended coverage report (instead of the 3D report) (because trees in the photos are blocking their technology from doing 3D in that case), then they cannot guarantee anything, but this report that I got was â€œ2D,â€ and itâ€™s only 4 s.f. off! So Iâ€™m pretty impressed! If they send you a 2D report, it comes as a 6:12 pitch roof, and you just go in and change the pitches of any roof planes to turn the report into 3D. Itâ€™s kind of neat just to play around with it and see how different pitches affect things. And their reports are extremely professional and clean looking â€“ they even tell you things like linear feet of ridges, hips, valleys, eaves, rakes, etc. Check them out. Another essential for the professional who has to repeatedly measure for a roof.

## 2. Calculating Starter, Hip/Ridge, Valley, and â€œWasteâ€ Shingles

I put the word â€œwasteâ€ in quotation marks because I heard of one homeowner that, hearing his roofer needed eight squares for â€œwaste,â€ offered to pick up those eight squares before the job, cut them all in half, and throw them in the bottom of the dumpster where they could get â€œhiddenâ€ by the rest of the trash of the re-roof job. No, no â€“ we donâ€™t mean â€œneedless waste,â€ we mean â€œhonest waste.â€ Some of that â€œwasteâ€ will be found under the shingles in your valleys, serving a very important and water-stopping purpose. Other parts of the â€œwasteâ€ are generated by the fact that hips and valleys must be cut on an angle, and you wonâ€™t be able to reuse the scrap piece easily or at all. It’s so important that you get this correct as you measure for a roof, because this is where things can really vary from roof to roof, depending on the architectural design of the roof.

• Starter

Well, letâ€™s start with the starter! Everybody puts starter on the eaves; some folks donâ€™t put it on the rakes, but I do, so thatâ€™s how weâ€™ll approach it. Itâ€™s good, I think, to put it on the rakes also, because it helps keep the overhanging shingles from sagging and looking sloppy, if nothing else. Starter is easy â€“ just measure the perimeter of your roof skin. Eaves + rakes. Then, finding that total length (for our example house, itâ€™s 317 feet), divide by 3â€² (length of one 3-tab shingle), and then by 26 (thereâ€™s 26 shingles per bundle, when dealing with 3-tab), and then by 3 again (three bundles to a â€œsquareâ€). In our case, we end up needing 1.35 Q, which weâ€™ll round to 1 2/3 Q (always round up to nearest 1/3, because youâ€™ll be ordering bundles, even though you talk â€œsquaresâ€).

$Starter = 1~2/3~Q~(of~3-Tab~Shingles)$

So 1 2/3 Q equals 5 bundles (just making sure you understand â€“ and Q is the symbol for a square, which is 100 s.f. of coverage area). Additionally, you could simplify your equation and just divide your linear feet measurement by 234 l.f. Moving right alongâ€¦

• Hip/Ridge

Letâ€™s talk hip and ridge shingles next. I donâ€™t know if other roofers buy the special hip/ridge shingle bundles, but everyone I know just uses 3-tab and cuts them into thirds. Now, the insurance people all throw around the equation L.F./84 for hip/ridge calculations, and that would be conservative, yet Iâ€™m not sure but that L.F./100 or even L.F./117 might not be more correct. It all depends on what your average exposure is â€“ do you nail your hip/ridge shingles with an exposure of 6â€³? Of 5â€³? Of 4.3â€³? Granting those averages (and those averages would have to include any waste along transitions, too) yields the numbers 117, 100, and 84 in the equation, respectively. So, I donâ€™t know for sure, but I think Iâ€™d average closer to 5â€³, if not even 6â€³, on my hip/ridge shingle reveal. But you can take your choice, I guess! On our example house, there was 144â€² of hip and 104â€² of ridge, yeilding a total of 248 L.F., which, after plugging into the middle equation, yields (248 L.F.)/(100 L.F./Q) = 2.48 Q (say) 2 2/3 Q. Again, thatâ€™s 8 bundles if you were wondering.

$Hip/Ridge = 2~2/3~Q~(of~3-Tab~Shingles)$
• Valley

Letâ€™s do valley shingles next. There are different options for shingling a valley (Open Valley, Woven Valley, Cut Valley, California-Cut Valley), but I prefer the California-Cut Valley (also known as â€œLong-Island Valleyâ€). Something about it just looks really professional. It seems to work better in preventing leaks. AND, itâ€™s the easiest one to do! So who wouldnâ€™t want to go for that? Long story short, if you do the California-Cut Valley, you need to account for the shingle that you lay onto the valley after the first side is complete which acts as the straight edge for the second side of shingles to pull from. So, just measure your valleys and divide by 39 3/8â€³ (length of an architect shingle) (or, if you want feet, by 3.281â€²), and then by 21 (21 shingles per bundle for architect shingles), and then by 3 (3 bundles per square). Or, putting all that together, just divide your valley lineal feet measurement by 207â€², and you should be good-to-go. For our example roof, we had 140â€² of valleys, which translates to 0.67 Q, but we better round up to 1 Q.

$Valley = 1~Q~(of~Architect~Shingles)$
• â€œWasteâ€

As you measure for a roof, you’ve got to consider for waste. Youâ€™ve probably heard, â€œHips and valleys eat up shingles.â€ Thatâ€™s true. If your house has a ton of hips and valleys, you need to figure for a LOT of waste. Think about it â€“ for every course of shingles that hits a valley, you lose a shingle to waste. Hereâ€™s how. In laying your California-Cut Valley (no, itâ€™s not much more wasteful than the others â€“ in fact, a woven valley will use about as many shingles), as you cross the center-line of your valley with the approaching (first) roof-plane of shingles, you are required to let HALF or more of your shingle go PAST the centerline, at the TOP of the shingles. So, for the first side, HALF or MORE of a shingle is lost for EVERY COURSE. Then, coming from the other side, as you do a special cut for the lay-on side of the valley, you waste UP TO half a shingle for every course. So, add those two up, and that means you waste AT LEAST one shingle for every course, and probably a little more. Weâ€™ll figure 1.15 shingles/course. Now, nobody wants to sit around and calculate how many courses of shingles every roof plane has! So, thereâ€™s an easier way. Just calculate your L.F. of valley, and use the table below to find what to divide your L.F. measurement by to know how many courses of shingles cross any valley over the course of the entire roof. It will also work for the hips, once we get there.

Finding Number of Courses of Shingles Crossing Any Valley for Entire Roof (Assumes a 5â€³ or 5 5/8â€³-Reveal in Your Shingle Installation [Standard]) (Wonâ€™t work for a switch-pitch, unless averaging two values)
Roof Pitch 3-Tab Shingles (5â€³ Reveal) Divide Total L.F. of Valley by This Number Architect Shingles (5 5/8â€³ Reveal) Divide Total L.F. of Valley by This Number
16:12 5 13/16â€³ 6 9/16â€³
14:12 5 15/16â€³ 6 11/16â€³
12:12 6 1/8â€³ 6 7/8â€³
11:12 6 3/16â€³ 7â€³
10:12 6 15/16â€³ 7 1/16â€³
9:12 6 3/8â€³ 7 3/16â€³
8:12 6 1/2â€³ 7 5/16â€³
7:12 6 5/8â€³ 7 7/16â€³
6:12 6 11/16â€³ 7 9/16â€³
5:12 6 13/16â€³ 7 5/8â€³
4:12 6 7/8â€³ 7 3/4â€³
3:12 6 15/16â€³ 7 13/16â€³
2:12 7â€³ 7 7/8â€³
1:12 7 1/16â€³ 7 15/16â€³

So, my example house (8:12 pitch) has 140 L.F. of valleys. Weâ€™re doing architect shingles (so, 5 5/8â€³ reveal). Therefore, 140 L.F./(7 15/16â€³) = 212 courses of shingles. Then, apply the waste factor of 1.15. WasteValley = (212)(1.15) = 244 â€œwasted valley shingles.â€ Divide by 21 (21 architect shingles/bundle) and then by 3 (3 bundles/square) yields 3.87 Q (say) 4 Q waste at valleys.

$Waste_{Valley} = 4~Q~(of~Architect~Shingles)$

By the way, donâ€™t get the feet and inches confused! A construction calculator sorts it out, but a scientific calculator would require CONSISTENT UNITS! Then thereâ€™s waste at the hips. Now, the hips arenâ€™t as bad. One side might eat UP TO half a shingle, and the other side probably eats UP TO a quarter, but weâ€™ll just say 1 total shingle/course wasted to be conservative. My example house had 144 L.F. of hips, so letâ€™s run back through the table with this figure.

$Waste_{Hips} = \frac{[(144 L.F.)/(7~15/16")](1.0)}{(21 shingles/pack)(3 packs/Q)} = 3.46~(say)~3~2/3~Q$
$Waste_{Hips} = 3~2/3~Q~(of~Architect~Shingles)$

What about rakes? Youâ€™ll probably waste at least 6â€³ every 2 courses, or 1 shingles every 12 courses. Divide your rake length by 5â€³ or 5 5/8â€³ (depending on whether youâ€™re doing a 3-tab or architect roof â€“ weâ€™re doing an architect in our example) to find # of courses, then divide by 12 to find number of wasted shingles due to the rakes. Then, divide by 21 shingles/pack and 3 packs/Q. In our case we had 103â€² of rakes:

$Waste_{Rakes} = \frac{(103')/(5~5/8")}{(12)(21~shingles/pack)(3~packs/Q)}$
$Waste_{Rakes} = 0.3~Q~(say)~1/3~Q~(of~Architect~Shingles)$

Then thereâ€™s the waste of the fact that 19 times out of 20, youâ€™ll have a partial bundle left, but you had to pay for the whole bundle, so figure 1/3 Q for both the architect and 3-tab shingles:

$Waste_{Partial Packs} = 1/3~Q~(Architect) ~and~ 1/3~Q~(3-Tab)$

Then again, thereâ€™s the waste of some of the shingles being damaged from the factory, some of the shingles being damaged during installation (such as dropping a shingles here or there, especially on steep roofs), etc. So figure 1 Q architect for that (for our size roof job):

$Waste_{Damaged} = 1~Q~(of~Architect~Shingles)$

Last of all, you want to have a little bit of a margin of safety. If you underbid a job, it eats up all your profit and then some really quickly! Figure 1 Q for a large job like this.

$Waste_{Safety-Factor} = 1~Q~(of~Architect~Shingles)$

So hereâ€™s all the waste together, for our example job:

$Waste_{Total} = 10~1/3~Q~(of~Architect~Shingles) ~and~ 1/3~Q~(of~3-Tab~Shingles)$

By the way, that adds up to a 18-20% waste factor for this roof (depending on whether you divide waste by geometric â€œskinâ€ of roof {=19.5%} or divide by [geometric-skin-of-roof + starter, hip/ridge, and valley] {=18%})! Of course, this roof is fairly cut-up, but it just goes to show you that there is math behind that waste percentage! So, summing up all the shingles for this roof:

$Geometric~"Skin"~of~Roof = 54~Q~(of~Architect~Shingles)$
$SHRVW~(Starter,~Hip,~Ridge,~Valley,~Waste) = 4~1/3~Q~(of~3-Tab~Shingles) ~and~ 11~1/3~(of~Architect~Shingles)$

Recapping by shingle type:

$Architect~Shingles = 65~1/3~Q$
$3-Tab~Shingles = 4~1/3~Q$

And finally, the grand total (this is what you bid your job by):

$Grand~Total = 69~2/3~Q$

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