AboutAzeem Hussain Expertise I can answer virtually any kind of question dealing with of Math 536 and below, my forte being in parabolic functions and analytical geometry.
I'm currently learning Linear Algebra, and cannot answer questions that deal with subject matter more advanced than that.
Experience I am neither a professor nor a teacher of this subject. I am merely a student who is gifted at mathematics and enjoys being of service to his community. I frequently tutor people in math and the results are usually great.
Publications Reflections, Riverside School Board (2005, 2006)
Education/Credentials Diploma of Secondary Studies from Chambly Academy High School, and IBO-MYP certificate as well. My lowest mark on a high school math final was 97%, peaking at 99% in 2006 and 2007 (second-highest Math 436 mark in the province). Being a Quebecer, I am fluent in English and French and can respond to questions easily in both languages.
Awards and Honors Pascal Math Competition, School Champion(2007)
How is the math done in the general example of the following specific situation?
You wish to cut down a sheet of plywood to 92-5/8 inches. You position the hook of your steel tape measure on the left end of the sheet at the bottom and you pull out the tape to the right so that it is everywhere flush with the bottom of the sheet's edge. You make a pencil mark at 92-5/8 inches. So far so good.
(Imagine that all of this takes place in the northeast quadrant of a grid.)
But what if the steel tape is hooked at the left end correctly but the right end -- the end where you're going to make a pencil mark at 92-5/8 inches -- is high by 1 inch? I can see that the resulting cut will be short, but I can't figure out by how much except by brute force. I'm hoping there's a formula.
The brute force method I've used is as follows:
Imagine a right triangle on a grid in which the lower left corner of the plywood is the origin. We know the hypotenuse is 92.625 inches (the intended length) and one of the legs is 1 inch (the error up).
Therefore, I think,
Other leg length = SQRT(92.625^2 - 1^2)
= SQRT(8579.390625 exactly)
= 92.619601732031 inches = actual x-value
Distance short = 92.625 - 92.619601732031
= 0.005398267969 inches
Have I got this right? If not, how does one calculate the distance short in this example?
And even if I do have this right, it doesn't do me much good because it's only one example. What if the hypotenuse (the intended length) is 96 or 60 or 12 inches instead of 92-5/8 and the error up is 2 or 3 or even 10 inches instead of 1 inch?
I'm looking for some sort of formula that expresses the distance short in terms of what we know about the right triangle I propose.
Is this idea better expressed in terms of angles? For example, if you're off by M degrees from horizontal, your measurement will be N inches short?
I think the two non-square angles formed in the example above are calculated like this:
Have I got this right? If not, what's the best way to calculate the small angle in this example? Isn't there something called an inverse of something, whether on my calculator or my Excel spreadsheet, that eradicates the 180/Pi factor? I did not take trig at all in school.
And even if I do have this right, it doesn't do me much good either, again because it's just one example.
Asked perhaps yet another way, as a radius of length 1 is swept counter-clockwise through a circle centered at the origin and starting at y=0, how does the x-axis value decrease as the y-value increases?
I want a formula that somehow takes the givens and turns them into the erroneous distance short, or maybe a percentage short, or something useful. The givens are the hypotenuse and the one (typically much shorter) square leg.
(If necessary we can also pretend that the novice volunteer can accurately estimate the angle of error, which provides us with a third given.)
How can I best express the relationships among the givens? Is there any way to do it for volunteers who don't have access to a device or table that does trig functions, or is that asking too much? Is there some way to graph the formula? Is there some way to graph several formulas, perhaps one for each of several selected distances or angles? This seems too complicated.
If you understand what I'm getting at, please let me know your thoughts. If you don't, can you suggest somewhere online where I should ask these math questions.
Thanks.
--Johnny
P.S. Just so you'll know, I've spent several hours trying to figure this out. If it helps, I can give you a URL to an Excel 2003 spreadsheet I've been playing with for a couple months now.
P.P.S. This is not for homework; it's for a how-to article I'm writing for the Kansas City, Missouri, affiliate of Habitat for Humanity that explains to novice volunteers how to measure and mark for various saw cuts on construction sites. The URL for the articles already in place, in case you still think this is homework, is http://barelybad.com/HabitatKC/. If you see I've made any mistakes, please let me know.
ANSWER: Hi Johnny,
You can breathe a sigh of relief; trigonometry can allow you to solve for everything you've mentioned above.
If the tape measure is a certain distance off from the bottom edge of the plywood sheet, the Pythagorean Theorem can be used to find the discrepancy (what you've done in your question is the perfect way). The formula c^2=a^2+b^2 can be used, where c is the hypotenuse, the intended length, a is the distance off the bottom (1 inch in your example) and b is the other leg. c-b will then be the distance short.
A single formula would be:
distance short=c-√(c^2-a^2)
Practically speaking, this is pretty much what you need. It's the simplest and most useful by far, and requires no trig operators. As for angles, I would strongly advise against any estimation, mainly because the angles are so small. (Again, your included example is correct. Look at the size of that angle.)
Nevertheless, here's a formula. Knowing the intended length c and the angle M (in DEGREES) from the horizontal:
distance short=c-c*cos(M)
Here's a crash course on angles. There are 2 main ways of representing them: degrees and radians. Degrees are what everybody knows of and uses practically. There are 360 degrees in a circle, 90 degrees in a right angle, etc. Radians are angle measures used in sciences and math, and are expressed in fractions of pi. These have little use outside. (You would tell someone to hold their saw at a forty-five-degree angle and not at a "pi over four"-radian angle.) Computers, as in Excel, tend to use radians, though. One degree equals 180/π radians, approximately 57 degrees (there are 2π radians in a circle). Excel has RADIAN and DEGREE functions for converting the two. If you're using the formula I've given you on a calculator, make sure it's set to degrees. In radians, it would be:
distance short =c-c*cos(Mπ/180)
For the record, inverse sine is the same as arcsine, and that's where you may have heard of that before.
As for your circle with a radius of 1, I don't see the relevance, but if you plot y=sin(x) and y=cos(x) on a graph in RADIANS, cos(x) will tell you the horizontal position and sin(x) will tell you the vertical position. The x-position will be the amount of radians rotated.
I hope you've noticed that with numbers such as the ones you've provided, the discrepancies are awfully small. You can graph the equations I've given you individually, where distance short is on the y-axis and a, or M for the second one, is on the x-axis.
Thanks for asking,
Azeem
---------- FOLLOW-UP ----------
QUESTION: Azeem, I am not as familiar with Allexperts as I could be, so please understand that I will rate your answer above as high as possible. I don't know whether to start a new question or ask a follow-up as I'm doing.
I would have asked you this in the Rate the Answer page, but I gather no one reads the comments.
If you want me to ask my follow-up in a separate question, for which you would get separate credit, just let me know.
I would have e-mailed you about this but I don't find an address for you.
Anyway, I do have a follow-up question or two, so let me know what to do next.
Thanks.
ANSWER: Hi Johnny,
If your question is an actual follow-up to your original question, ask it as a follow-up. If it is a separate question that isn't directly related to the same situations about which you had originally asked, then ask a separate question.
As far as I know, credit for a follow-up is treated as credit for an additional question. I really don't mind, but thanks for being considerate.
Azeem
---------- FOLLOW-UP ----------
QUESTION: March 9, 2009
With regard to your earlier answer, you'll remember that I am interested in the math of errors when a steel tape measure is mistakenly tilted at various distances and angles from horizontal.
I want to write in Excel 2003 two little calculators. Each requires two givens, one of which is the intended distance = hypotenuse = c. The first calculator takes as its other given the DISTANCE_high the tape measure is mistakenly set to. The second calculator takes as one given the ANGLE_high the tape measure is mistakenly set to and, again, the other given is the intended distance = hypotenuse = c.
The result in both calculators is the distance_short the measurement will be. I think I've got the first calculator figured out. As a bonus, I also show the resulting angle of error, which I think is
90 - (ACOS(distance_high/c) * (180/Pi) ).
Let me know if I got that wrong.
But as to the second calculator, where the given is the angle_high, how do I translate that into the mistaken distance_high the tape measure must be?
For the distance_short formula I did use
c - (c * COS(angle_high / (180/Pi) )).
If that is right, then let me ask my real question. How do I formulate the distance_high the tape measure must be given the angle_high and the hypotenuse c?
For example, if your tape measure is 1 inch high over a distance of only 2.125 inches, the measurement will be short by exactly 1/4 inch, and the resulting angle of error will be over 28.072 degrees. This surprised me, and I hope you can confirm that those answers are accurate.
Assuming I got this right so far, what is the formula for translating the angle_high and the hypotenuse c into the mistaken distance_high? I realize this might involve more algebra than trigonometry, but I've already burned up a lot more brain cells on this problem than I can afford, and I last took algebra during an early Trudeau administration.
This formula is correct, but can be simplified as:
angle_of_error=ASIN(RADIANS(distance_high/c))
Your distance short formula is also correct but can be simplified as:
distance_short=c-c*COS(RADIANS(angle_high))
The formula you seek for the distance high is the following:
distance_high=c*SIN(RADIANS(angle_high))
The situation you've given has all the correct values. Try it out for yourself if you're surprised. The smaller the intended length, the bigger difference the height/angle high will make.
