If you know the lengths of the two square legs of a right triangle (which means you know the slope), is there a formula that tells you the x-axis value directly under a given distance measured along the hypotenuse? I’m afraid the answer is too easy, but after MANY hours of noodling I still can’t figure it out. I hope you, a SD expert, can.
My example is pulling out a steel tape measure from the bottom left of a sheet of 96-inch plywood all the way to the right, past the right end of the sheet, except that you mistakenly raise the tape by exactly 1 inch at the 96-inch point, i.e., the right side of the sheet. Then you find the 72-inch mark on the tape, which is the hypotenuse of a slim triangle, and you make a mark on the sheet. That mark will be short of 72 inches by some amount or other as measured along the purely horizontal x-axis, and my question is, How do I calculate that value on the x-axis that is directly below the 72-inch value on the hypotenuse?
As I see it, any formula must incorporate these three variables:
(1) The distance the tape was mistakenly held too high (the Rise part of Rise over Run) = 1 inch in the example
(2) The length of the sheet from left to right (the Run part of Rise over Run) = 96 inches in the example
(3) The point along the hypotenuse of that triangle where you (mistakenly) make your mark = 72 inches in the example
The result of the formula will be a number less than (3) by some amount. What formula produces this answer given the three variables?
I want to be able to say, "At 96 inches if your raise your tape 1 inch too high, your mark at 72 inches on the tape will be actually be short by X inches along the bottom. Or “At 92-5/8 inches if you raise your tape 3-1/2 inches too high, your mark at 60 inches on the tape will actually be short by Y inches along the bottom.”
You get the idea, I hope.
Thanks for any suggestions you can offer to get my brain moving in the right direction again.
If you know the lengths of the two square legs, “mere Pythagoras” will get you the length of the hypotenuse. From there you can use proportions, since drawing a line from a point on the hypotenuse to the point directly below it on the x-axis will give you a similar triangle to the one you started with.
If I understand you properly (I’m not sure I do), then you measure a distance L0 along the horizontal from the origin at the lower left corner of the sheet. Only you don’t actually run the tape directly across the sheet – it’s a distance a too high, but goes all the way to the end of the sheet. If you drew a line along that tape measure to the other side of the sheet, it would be a length given by the square root of L0[sup]2[/sup] + a[sup]2[/sup].
Now you measure a line a distance L1 along that tilted line, and you want to know what point on the edge (the x-axis) below corresponds to that point. Since the triangle you get by dropping the vertical down to that edge is similar to the one you first drew, that went all the way out to the edge, the proportions will be identical, and if L1’ is the distance along the edge, you’ll have L1’/L1 = L0/SQRT(Lo[sup]2[/sup] + a[sup]2[/sup])
In other words, L1’ = L1*L0/SQRT(L0[sup]2[/sup] + a[sup]2[/sup])
For your example, with L0 = 96, a = 1, and L1 = 72, you get L1’ = 71.996 inches
If a << L0, as in your case, then a useful approximation is
Use the sine and cosine. If length along the hypotenuse is L, then the horizontal distance and vertical distance are Lcos(theta) and Lsin(theta) respectively. Where theta = inverse tangent of the ratio of the legs of the original triangle.
beowulff and Hector St Clare and md2000, thank you for presenting the trig solutions. I instituted both Hector’s and md2000’s in a spreadsheet to see, and they match exactly, so I’m pretty sure the formulas I created are right.
And they exactly match CalMeacham’s results, which use Pythagoras only, which further proves the all the formulas.
CalMeacham, thanks you for such a helpful answer, and may I ask how you arrived at your approximation, which divides by 2? It is remarkably accurate.
It’s a very common and straightforward approximation . The square root of 1 + E is very nearly 1 + E/2 if E << 1.
They’re the first two terms of the Taylor expansion of the square root. Alternatively, if you square 1 + E/2 you get 1 + E + E[sup]2[/sup]/4, and if E <<1 you can ignore that last term as negligible.
It’s a trick known as the “binomial expansion”. Basically, if you have a quantity like
(1 + x)[sup]n[/sup],
you can expand it out as a polynomial in x:
1 + n x + n(n-1) x[sup]2[/sup]/2 + …
If x is a small number, then you can throw out all of the terms except the first two. For example, if x is 1/1000, then x[sup]2[/sup] is 1/1000000, and so just using the first two terms gives you an answer that’s correct to within a few parts in a million.
since 1/√x = (x[sup]1/2[/sup])[sup]-1[/sup] = x[sup]-1/2[/sup]. From there, applying the binomial expansion above (with n = -1/2) gives you CalMeacham’s formula.
Your mark at 72" is 3/4 of the length, so the perpendicular from there will be 3/4".
This means we have a right triangle with a hypotenuse of 72, so the base can be calculated by subtracting the square of 3/4 (9/16) from the square of 72 (5184) and finding the square root of the result.
5184 - 9/16 = 5183 and 7/16
The square root of 5183 and 7/16 is 71.996093644 (using Google)
So the error is 0.0039, or about 1⁄250.
Since you are using a cheap steel tape made in China; you don’t have any idea how it may have stretched or expanded in the hot workshop, and your pencil is probably wider than 1/250", the error is insignificant.
For a general rule, you might do best to create a graph.
Of course, you can also continue with more and more terms to get to whatever degree of accuracy you wish. When x is small though, you can get very close to your answer very quickly (i.e. the series ‘converges’ quickly). The binomial series really is incredibly useful.
There’s one formula I can’t figure out using trig, which is where you know the hypotenuse and one of the non-square angles. What formulas calculate the lengths of the legs?
For example, if the hypotenuse is 96.0052081920559 inches and one of the angles is 0.596809451229177 degrees, how does one arrive at one leg length of 96 inches and the other leg length of 1 inch?
I’ve spent as much time as I care to – about 45 minutes now – trying to figure this out. I learned and forgot trig in high school, and I don’t want to have to learn it again.
If you call the hypotenuse r and one of the angles theta (I can’t figure out Greek letters here), then the legs are rcos(theta) and rsin(theta) respectively.
If you mean R of 96.0052081920559 times the cosine of the known angle of 0.596809451229177, I arrive at 53.9555347568169 inches. If you mean R of 96.0052081920559 times the sine of of the known angle of 0.596809451229177, I arrive at 79.4090691867492 inches.
In fact, those are the formulas I had, and I recently assumed they were wrong somehow because they produce unexpected results. I expected a triangle whose square legs equaled 1 inch of Rise over 96 inches of Run.