# 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. (≠ Amount of Shingles Needed)

Before we show how to calculate the geometry of a roof’s surface, let’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 you’re finished getting all the particular shingle elements, you may have 52 squares (5,200 s.f.) total. But we’ll get to that later on. And are we on the same page that 1 “square” (abbreviated “Q”) = 100 square feet? Okay, just wanted to make sure. Let’s dive in! Forgotten your high school/college geometry? Let’s go over it first.

• “Roof Geometry Placement Test”

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, it’s “x” (eave length) times “y” (rake length). Since there’s two sides, you multiply that by 2. For a simple hip, to find the trapezoids, you find your average measurement of “x1″ (long eave length) and “x3″ (ridge length), then multiply that by “y” [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 “x2″ (small eave length) times “y” and divide by two. Since there are two trapezoids and two triangles, it’s simple math to get the grand total. Take one last look at this “pre-test,” and if you’ve got all that, we’ll “place” you in “MAT-361-R Euclidean Geometric Concepts for Roofs“!

”Geometry Roof Placement Test”

Did you know? In Britain, “trapezoids” and “trapeziums” refer to the opposite things that they do in the US? Check it out – we’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.
• MAT-361-R Euclidean Geometric Concepts for Roofs”

No, you don’t have to know what “Euclidean” means, but we’ll give you that as a bonus. “Euclidean” 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 “class,” we’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 we’ll provide a computerized version in a moment – 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 it’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.

Hand-Drawn Roof Plan
13 Roof Planes on This Roof
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 (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 we’ll just neglect that). So let’s jump right in with roof plane #1. We’ll be looking at the math behind each roof plane.

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 let’s divide it into three sub-regions. First, an exact rectangle on the left (28′ x 11’6); then, a trapezoid with a horizontal “altitude” of “?”, and two vertical “bases” (one is clearly 24′, and the other one, though unmarked, is obviously the same as the 11’6 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, there’s a triangle with a “base” of “??” and a height of 24′. So, we just need to find what the values “?” and “??” are. We can agree to the following, right:

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

Further, because we have two right triangles (one has hypotenuse 16’6 and the other has hypotenuse 31’6), and because we know that their two furthest angles will be the same [but NOT 45°!]), 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 “?” 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 “??” acquired, go back to Equation 2a, and solve for “?”.

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

Couldn’t we have measured for ? and ??, instead of doing the math? Yes, but sometimes that’s easier said than done. What if you don’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 can’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. It’s not that hard. You could also have solved for “??” in this way, using a construction calculator and pressing the following buttons:

2 4 Feet Diag 8 Inch Pitch Run

Basically what you’re doing there is telling the calculator that you have a 24′ diagonal, an 8:12 pitch, and then by pressing the last button “Run,” you’re asking for the length of the run, which it returns to you as 19′-11 5/8″. By geometry, THAT “Run” will happen to be equal to our “??” figure (to see that visually, skip down to the large hand-sketch of the pyramid). Now, you say, “Wait a minute! Didn’t we find our unknown, ‘??’ to be 21′?” You’re right – we did. The problem is that my 60′ measurement was too conservative (it’s hard to measure a long length like that with one person) and probably should have been 59′ or maybe just a wee less than that. Then, when you reiterate the formulas above, you get something like 20’4, which is only about 4″ off from our construction-calculator (and probably more accurate) measurement. When you’re in the field, you do your best, but some calculations are going to have to be “rounded up.” Anyway, I just used 21′ in the area calculations below.

Two Ways to Find “Eave Run” Below a Hip – Using Slant Height & Roof Pitch; Or, Using Hip Length & Hip Pitch (“Triangle-2″ Type)

Now that we’ve got “?” and “??,” 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 “exercise,” 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 on… Were you tempted to think that “??” would have been 24′, the same as the other leg, since it’s a right triangle and the other angles looked to be 45°? 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 you’re looking at the bird’s-eye-view diagram like it’s 2D. In reality, the roof is a 3D structure, and what is 45° on the bird’s-eye view actually happens to be 48.8° 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 – just I’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 it’s triangle, the hip’s “pitch” 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.

Three Geometrically Separate Triangles to Keep Straight in Roofing

So let’s look into why our previous unknown “??” could not be equal to 24′ (the “slant height”). Consider first that the two angles of the triangle, while looking the same, are quite different. Here’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 50′ x 50′ building, but it’s difficult to draw the exact measurements (at least for me). Still, you can see the gist of what I was getting at – in the 2D “bird’s-eye” view, ∠a and ∠b appear to be identical (i.e., 45° each); however, in the 3D view, it’s obvious that ∠a is much smaller than ∠b, and as you approach a perfectly-vertical, infinitely-high spire, ∠a would approach 0°, and ∠b would approach 90°, while STILL appearing to both be equal (45°) in the 2D “bird’s eye” diagram! That’s why roofers have to be careful to remember what PLANE they are in, and not get the 2D and 3D aspects mixed up!

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, I’m sure you can get past the snafus I’m running into. You can check it out at www.chiefarchitect.com. But since I wasn’t able to use Chief to create what I wanted, here’s the hand sketch. It better illustrates the idea of just how different an angle can look in 2D versus 3D.

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

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

More Than One Triangle in a Pyramid!

Note that triangles 1 and 4 have the same measurements (unless you have a “switch pitch roof,” where the front-back pitch is one thing, and the left-right pitch is (usually) steeper). Notice that there’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 “insanely steep” roof, the pitch is 240/12. However, if you were to take the right half of “triangle 2″ out of our “insane” roof, and stand it upright (it already is almost vertical), the “pitch” of the “hip” (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 hip’s RELATIVE pitch, stood-upright, was about 15:12 (a HUGE change from 8:12). The point is, unless you’re dealing with insanely steep roofs, the hip’s “two” 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 “walk” right along it’s edge), it would actually be 170:12 (SIGNIFICANTLY LESS – but still not safe to walk on!). How did I get that? Well, in the base of the pyramid, it’s 50′ both ways to the center point. So, just use the Pythagorean Theorem to find that base E = 70′ 8 1/2″. Then, the height for that triangle is 1000′. Revisit Pythagoras and you get a pitch of roughly 170:12. That’s why it’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? It’s because for a hip, you’re actually standing on roof planes that are trying to throw you off away from the hip; whereas with a valley, you’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 “hand-rails” (the two roof planes ascending out of the valley) to protect you further. You practically can’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, you’re really just climbing a 10:12 with the added comfort of something to sort of hang on to (the hip) – 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 you’re sitting on it and scootching up, then it’s a 10:12 with a somewhat-handrail, not a 7:12 (which is, geometrically, the pitch of that imaginary line, the “hip” [remember, it’s really a LINE, not a PLANE]). By the way, here’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.

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

Notice that each time you add two more inches to the “rise,” 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. It’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 “feet-and-inches” mindset and keep perfect accuracy. It also has buttons for rise, run, diagonal, pitch, etc. It’s a must-have for roofers. It’s extremely user-friendly and I’ve never been frustrated with it. Somebody thought before making that calculator – you can’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. It’s even better than the handheld calculator – and you’ll have it with you wherever you are! An absolutely essential tool if you’ve got to measure for a roof!

An Excellent Construction Calculator

Okay, back to calculating our roof planes.

Geometry of Roof Plane #2

Well, this roof plane is easy, right? It’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 we’ll neglect that due to the difficulty with Chief. Sometimes on the field you will have to kind of “estimate” a small area that’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 we’re getting here is really close.

Geometry of Roof Planes #3 and #4

So for roof plane #3, it’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, it’s what? That’s right – a trapezium! I don’t know any formula for that, so we’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 – ?, ??, and ??? – but I’m fairly confident we can find them all. I’m going to borrow some math from roof plane #3 to help us get variable “??.” 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 “??”:

2 2 Feet 6 Inch Diag 3 Inch Pitch Run

That spits out ?? = 21′-9 15/16″.

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

Then we can write equation 4 to solve for “?”:

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

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%

2 Feet 2 Inch 1 / 1 6 Conv

x2%

=

1.532959 SQUARE FEET
Conv

√xClear
1′-2 7/8″

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 (24′) and (12′), we would have come up with 144 s.f., which wouldn’t have been very off. I’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 along…

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

Notice that I’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, I’ve labeled our seven unknowns as lower case a through g. First point. a≠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 you’re looking at a bird’s-eye (plan) view of a roof, leg a of a right triangle will NEVER equal leg b! Okay, back to our math – 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 you’re probably wondering where I came up with a 14.5:12 pitch, right? Well, remember that triangle A is a “type 2″ 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 “type 2″? Because that’s the triangle that forms the surface of a pyramid – the other two (or three) triangles had to do with the INTERIOR of the pyramid. But our triangle “A” is on the surface of the pyramidal shape, so we have to use the values for “type 2″ triangle. In fact, ANY TIME you only know the HIP length (and don’t know the SLANT HEIGHT) and you need to find the length of “run” that goes along the eave line (or along the pitch-transition-line, as here), you need to know these “triangle 2″ pitches (so I’d recommend writing them down – I’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 12′) *Hip Length (for a RUN value of 12′) *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 20′ 0″ 23′ 3 7/8″ 20 : 12 11 5/16 : 12
14:12 18′ 5 1/4″ 22′ 0″ 18 7/16 : 12 9 7/8 : 12
12:12 16′ 11 5/8″ 20′ 9 3/8″ 17 : 12 8 1/2 : 12
11:12 16′ 3 3/8″ 20′ 2 11/16″ 16 5/16 : 12 7 3/4 : 12
10:12 15′ 7 7/16″ 19′ 3 3/8″ 15 5/8 : 12 7 1/16 : 12
9:12 15′ 0″ 19′ 2 1/2″ 15 : 12 6 3/8 : 12
8:12 14′ 5 1/16″ 18′ 9 1/8″ 14 7/16 : 12 5 11/16 : 12
7:12 13′ 10 11/16″ 18′ 4 1/4″ 13 7/8 : 12 4 15/16 : 12
6:12 13′ 5″ 18′ 0″ 13 7/16 : 12 4 1/4 : 12
5:12 13′ 0″ 17′ 8 5/16″ 13 : 12 3 9/16 : 12
4:12 12′ 7 13/16″ 17′ 5 1/4″ 12 5/8 : 12 2 13/16 : 12
3:12 12′ 4 7/16″ 17′ 2 13/16″ 12 3/8 : 12 2 1/8 : 12
2:12 12′ 2″ 17′ 1 1/16″ 12 3/16 : 12 1 7/16 : 12
1:12 12′ 0 1/2″ 17′ 0″ 12 1/16 : 12 0 11/16 : 12

Once you understand that, it’s easy – “a” is simply the “rise,” and “b” is simply the “run.” So, with those values found, the area for triangle “A” 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 “a,” we can find “c”:

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

Then, g = b – (2’6″ converted-to-run-value). So, plug 2’6″ through the exact same process as 18’8″ to arrive at a “run” value of 1′ 7 1/8″. 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 “d” is simply the hip measurement (2’8″) converted to a “run” 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 “run” value for a 2’6 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 “funny” operation on hip “16’6″ to convert it to the relative “run” 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:

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

Similar to roof plane #1, this time it should go lots quicker! Notice that the left rake (13’6) is smaller than the right one (15’3). 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 “a” and “b” of trapezoids B and C, we cannot use the slant height (because the hips don’t go all the way down – it would provide a value almost twice as large as what we need); but, we can use the two hip lengths with their pitches (“triangle 2 type”). 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 “a” and “b” would have been 3/8″ different (in error), otherwise. Now, let’s find the area of trapezoids B and C, and the total area of roof plane #6:

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

Don’t you love large katy-wompus roof planes like this one? Not infrequently, as you measure for a roof, you’ll run into them. They’re critical to get right, yet how do you do it?? Well, we’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 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. I’m going to split it up into six geometric shapes (A through F). Then, we have ten unknowns to solve for (a through j). Let’s find “a” and “b” 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 “d” is a little tricky – “d” will be equal to “a” minus the “RUN” value of the 9′-hip (using “triangle 2″ value). First find the funny “RUN” 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:

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

Let’s skip “f” 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. We’ll use the hip length and apply the “funny” (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,”f” 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 “j” is just the ridge length minus “e”:

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

Let’s solve for our areas, and then for the total area of this hexagonal roof plane:

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

On this roof plane, c = 2’8″ (look at the hip measurement above). Also, by geometry, b = c = 2’8″. 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 they’d both still be equal. It’s not the same situation as we had earlier. Finding a is also easy:

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

There’s one variable I forgot – we’ll call it “d” and it would be a horizontal line that would strike through the letter capital C. It’s the “slant height” of triangle C, and to find it, we imagine that line b in our drawing is a hip, and apply that “funny” operation again:

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

The reason we can’t say A = 1/2(b)(c) is because the angle between them is not 90°. It is 90° in 2D, but as you lift that point higher and higher, the angle becomes smaller and smaller. I’m not sure what the angle is at our 8:12 pitch, but the point is, we can’t use those legs. We have to use the 4′ 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 one’s easy – it’s a parallelogram! Just base times height to find the area. The only catch is, make sure you use the correct “height.” 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 6’6 as the “height,” you’ll get a wrong answer. The height we want is 4’6:

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

Hopefully by this point, you’re getting the feel for things. I’m dividing roof plane #12 into two triangles, whilst dividing roof plane #13 into a rectangle and a triangle. We’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 it’s angles, and you should understand what I’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 hadn’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 = 12′ 7~7/16″
$A_{12} = \frac{1}{2}*(8')*(9' 6") + \frac{1}{2}*(2' 6")*(3' 0~1/16") = 38 + 8 = 46 s.f.$
$A_{13} = (7' 6")*(15' 3") + \frac{1}{2}*(10' 6")*(12' 7~7/16") = 114 + 133 = 247 s.f.$

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, we’ll say that you successfully passed MAT-361-R Euclidean Geometric Concepts for Roofs”!

By the way, you can order roof plans almost overnight from “Eagle View Technologies” 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! I’d highly recommend them! If you’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. You’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 “2D,” and it’s only 4 s.f. off! So I’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. It’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 – 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 “Waste” Shingles

I put the word “waste” in quotation marks because I heard of one homeowner that, hearing his roofer needed eight squares for “waste,” 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 “hidden” by the rest of the trash of the re-roof job. No, no – we don’t mean “needless waste,” we mean “honest waste.” Some of that “waste” will be found under the shingles in your valleys, serving a very important and water-stopping purpose. Other parts of the “waste” are generated by the fact that hips and valleys must be cut on an angle, and you won’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, let’s start with the starter! Everybody puts starter on the eaves; some folks don’t put it on the rakes, but I do, so that’s how we’ll approach it. It’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 – just measure the perimeter of your roof skin. Eaves + rakes. Then, finding that total length (for our example house, it’s 317 feet), divide by 3′ (length of one 3-tab shingle), and then by 26 (there’s 26 shingles per bundle, when dealing with 3-tab), and then by 3 again (three bundles to a “square”). In our case, we end up needing 1.35 Q, which we’ll round to 1 2/3 Q (always round up to nearest 1/3, because you’ll be ordering bundles, even though you talk “squares”).

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

So 1 2/3 Q equals 5 bundles (just making sure you understand – 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 along…

• Hip/Ridge

Let’s talk hip and ridge shingles next. I don’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 I’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 – do you nail your hip/ridge shingles with an exposure of 6″? Of 5″? Of 4.3″? 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 don’t know for sure, but I think I’d average closer to 5″, if not even 6″, on my hip/ridge shingle reveal. But you can take your choice, I guess! On our example house, there was 144′ of hip and 104′ 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, that’s 8 bundles if you were wondering.

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

Let’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 “Long-Island Valley”). Something about it just looks really professional. It seems to work better in preventing leaks. AND, it’s the easiest one to do! So who wouldn’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 3/8″ (length of an architect shingle) (or, if you want feet, by 3.281′), 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 207′, and you should be good-to-go. For our example roof, we had 140′ of valleys, which translates to 0.67 Q, but we better round up to 1 Q.

$Valley = 1~Q~(of~Architect~Shingles)$
• “Waste”

As you measure for a roof, you’ve got to consider for waste. You’ve probably heard, “Hips and valleys eat up shingles.” That’s true. If your house has a ton of hips and valleys, you need to figure for a LOT of waste. Think about it – for every course of shingles that hits a valley, you lose a shingle to waste. Here’s how. In laying your California-Cut Valley (no, it’s not much more wasteful than the others – 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. We’ll figure 1.15 shingles/course. Now, nobody wants to sit around and calculate how many courses of shingles every roof plane has! So, there’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 5″ or 5 5/8″-Reveal in Your Shingle Installation [Standard]) (Won’t work for a switch-pitch, unless averaging two values)
Roof Pitch 3-Tab Shingles (5″ Reveal) Divide Total L.F. of Valley by This Number Architect Shingles (5 5/8″ Reveal) Divide Total L.F. of Valley by This Number
16:12 5 13/16″ 6 9/16″
14:12 5 15/16″ 6 11/16″
12:12 6 1/8″ 6 7/8″
11:12 6 3/16″ 7″
10:12 6 15/16″ 7 1/16″
9:12 6 3/8″ 7 3/16″
8:12 6 1/2″ 7 5/16″
7:12 6 5/8″ 7 7/16″
6:12 6 11/16″ 7 9/16″
5:12 6 13/16″ 7 5/8″
4:12 6 7/8″ 7 3/4″
3:12 6 15/16″ 7 13/16″
2:12 7″ 7 7/8″
1:12 7 1/16″ 7 15/16″

So, my example house (8:12 pitch) has 140 L.F. of valleys. We’re doing architect shingles (so, 5 5/8″ reveal). Therefore, 140 L.F./(7 15/16″) = 212 courses of shingles. Then, apply the waste factor of 1.15. WasteValley = (212)(1.15) = 244 “wasted valley shingles.” 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, don’t get the feet and inches confused! A construction calculator sorts it out, but a scientific calculator would require CONSISTENT UNITS! Then there’s waste at the hips. Now, the hips aren’t as bad. One side might eat UP TO half a shingle, and the other side probably eats UP TO a quarter, but we’ll just say 1 total shingle/course wasted to be conservative. My example house had 144 L.F. of hips, so let’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? You’ll probably waste at least 6″ every 2 courses, or 1 shingles every 12 courses. Divide your rake length by 5″ or 5 5/8″ (depending on whether you’re doing a 3-tab or architect roof – we’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 103′ 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 there’s the waste of the fact that 19 times out of 20, you’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, there’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 here’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 “skin” 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):

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