Steps and Stringers - February 26 - March 1, 1996
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Last spring I moved some steps on my deck and made them much fancier - they used to go down along a wall in a corner, but now they come out at an angle from the deck and the house and get larger as they go down. The idea is that they are much more inviting, not to mention a lot easier to negotiate with an armful of groceries. :-)Steps are usually built on two pieces of wood called "stringers." These are the supports, and are in this case two 2x10-inch boards with notches cut out of them for the steps to sit on. The tricky part in cutting these stringers was to find out at what angle they met the top board they would be attached to - we bought the stringers pre-cut at the lumber yard, so we wouldn't have to cut all the notches. Here is a top view of the steps, which go out 42 inches from the deck (that's horizontal distance).
40.5" _ __________ | * * | * * | * * * * 42" * * * * | * * | * * | * * _ *____________________________* 5'So the question was, at what angle should the stringers (represented by the *s) meet the top board, which is the one that is 40.5"? If we cut them at 90 degrees, they wouldn't lie flat against the header. I needed to cut them with the circular saw, and it has a little guide on it with degree markings, so I just rounded to the nearest degree and figured that would be close enough.This was one of the few times I have used trig in a completely practical manner and the person helping me was impressed!
Annie says: Solutions
This problem made perfect sense - if you were me! There was some confusion at first in the statement of the problem, for which I apologize, and I left out a measurement. Once that was straightened out, enough people got it right that I know it wasn't totally off the wall.
This problem would be a lot easier to understand standing in my back yard with a circular saw and a bunch of pressure-treated wood. I think my narrative skills broke down in relating the experience.
One interesting thing about these solutions. You'll notice that there are several different answers - 103 degrees, 77 degrees, and 13 degrees. They are really all the same thing; it just depends on which angle you are looking at. It would make more sense, again, if you had the circular saw in your hands - it only goes from 0 to 45 degrees anyway, so 13 was the answer I was looking for in this case.
See if the solutions below help you understand the problem any better. I'm just digging myself into a hole if I talk about it any more. :-)
Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are also available.
Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California A B ---------- *| |* * | | * * | | * * | | * * | | * * | | * * | | * ------------------------ C E F D angle ACD = angle BDC = 76.9 degrees angle CAB = angle ABD = 103.1 degrees If I draw AE and BF perpendicular to CD as shown above, then I could solve angle ACE and angle CAE by using trigonometry. I know AB = EF = 40.5" and CD = 5' = 60". I also assume EF sits in the middle of CD, then CE = FD = (60-40.5)/2 = 9.75". AE = BF = 42" were given. I then solve the trigonometric function tan ACE = 42/9.75 for angle ACE, which is 76.9 degrees. angle CAB = 180 - angle ACE = 103.1 degrees.
Jeff Lavallee Grade: 12 School: Croton-Harmon High School The angle between the stringers and the top board is 103 degrees, rounded to the nearest degree. I arrived at this solution like this: 5' is 5*12 or 60 inches. The length of the bottom side of the triangle formed by the stringers, the bottom board, and the altitude of the trapezoid is 1/2 the difference between the lengths of the top board and the bottom board, which is 1/2 * (60-40.5) = 9.75. The angle is the Arc Tan of the length of the bottom of the triangle over the altitude of the trapezoid, which is Arc Tan Theta = 9.75/42 = 13 degrees (rounded to the nearest degree). Adding this to 90 degrees gives us the measure of the angle between the stringers and the top board. 40.5" _ __________ | *|90 * | Theta=13--> | * | * | * * | * 42" * | * * | Altitude * | * | of * | * | Trapezoid * | * | * _ *_________|__________________* 9.75 5' ^ | Bottom of triangle formed by stringers and altitude
Selena Bradbrook Loreto College, Marryatville Adelaide, South Australia. The answer is 13 degrees. This is because segment AB is 42 inches long. Given a segment BC is 9.75 inches long. This is because 5' = 60" ( 5 x 12 ) 60 - 40.5 = 19.5 half of 19.5 is 9.75. I halved this because a trapezium (the shape of your deck) is made up of two triangles and a rectangle. I need the base of one triangle, thus I halve it). Angle ABC is 90 degrees Tan q = opp adj = 9.75 42 ( tan -1 ) = 13.0 Therefore the answer to this problem is 13 degrees. The other Angle = 77 degrees
Catherine Dinham Year 10 Loreto College, Marryatville Adelaide, South Australia. When we first started this problem we were unsure how we were going to solve it because we believed we needed the length of the steps to work it out. We also became confused with what you were saying when you said that the steps come out from the house at a 45 degree angle. We guessed that it was going down from the house at 45 degrees. Here's what I did: We need to find the Base angle. Tan x = Opp Adj Tan x = 42 9.75 Tan x = 4.30 x = 77 Therefore because of alternate angles we can see that the angle at which the wood needs to be cut is 77 degrees.
Brian Gordon Dartmouth '92 I figure you should miter-cut the sides at 103.07 deg. Drop the perpendiculars from the top step to the bottom to give you a rectangle that's 40.5 x 42 inches. This leaves 19.5 inches left on the bottom step, divided into two 9.75 sections (I assume you're making the steps perpendicular to the plane of the back of the house). Then the angle at the top of the steps will be 90 (from the rectangle) + the arctangent of 9.75/42. That's 13.07 degrees, so the total angle is 103.07. I was watching the Old Yankee Workshop on PBS, and Norm had to cut something at a nine degree angle. He said that he figured that to be 1 7/8 inches of rise for every foot of run. Actually it's closer to 1.9 inches, but I thought it was neat that he was using some trig himself. :) --bri
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