Application of algebra

And trig. The basic 3 functions. A few years ago I built a trolley to lift roofing shingles on a ladder:

/ / / / / / / / / + / ---- | / x \-------O/ ---- \ / \ / \ / O/ / / / /

There was an electric hoist at the top & the big question was where to attach the cable to the trolley, so as to have equal weight on the wheels. Too low and the cable would lift the top wheels; too high & it would lift the bottom wheels. Somewhere in the middle would be just right.

I solved it by summing moments to zero around 2 points, solving simultaneous equations with lots of sin, cos, & tan values. (If you're thinking "Why simultaneous equations when there's only 1 unknown?", you're right, but for some forgotten reason it was more convenient to have 2 or 3 unknowns.)

Bob

Thats OK as long as it DOSEN'T have an = sign on it.:-) ...lew...

No, he was using algebra. No higher math at all.

Teaches you to think abstractly, which is an extraordinarily valuable skill even if you never take another integral.

Exactly. Generally the people who can find no use for mathematics are the ones who never took the time to learn any mathematics to use.

Regards, Marv

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I'm horrible at higher math, but even I know you split that into a horizontal cylinder and a sphere (for the two end shells). Solve the height/volume for each separately, and add the results together.

And don't forget to allow for a striking pad in the bottom of the tank at the stick hole - underground storage tanks place a big piece of 1/2" plate there so you don't drop the stick hard a few thousand times and manage to crack the inner wall. Or just 'the wall' on an old single-wall tank.

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While tanks with hemispherical ends exist, most tanks have domed ends that are not hemispheres. Typically, they're a portion of a spherical surface that has a center somewhere in the interior of the tank.

Given only the OAL of the tank and the length and diameter of the cylindrical part of the tank, the math becomes an interesting exercise.

Bruce you are right, the equation for the volume of a partially filled horizontal cylinder as a function of the height needs to be derived as well as the partially filled domes on the end. Of course a straight forward, less exotic way is to fill the tank a gallon at a time and mark your stick.

Stu

Reminds me of a story my 6th grade math teacher (back in the 60s) told about Thomas Edison... don't have any idea as to whether it's true or not.

Edison hired a mathematician to calculate the volume of a light bulb. After two weeks, the mathematician hadn't quite come up with the answer. So Edison filled the bulb with water, poured the contents into a beaker and checked the scale. Can't say it was that accurate, but still a nice story.

-Bruno

You know Marv that would make a good "Sig line" if I were into such. (like a few on here :-) ) ...lew...

How about: Constipated mathematicians work it out with a pencil! ;-)

Actually it was probably more accurate, as getting and using enough dimensions to characterize what was probably not a simple or regular shape would be a real pain.

A^2 +B^2 -C^2 = 0

looks like algebra

That's not algebra. That's geometry.

That is Algebra.

Geometry is the angle that distends B is a , that distends A is b. When the length of A=B then a=b and the angle of a or b = 45 degrees.....

There is(are) a trig version(s)

There are other math versions as well.

Martin

Mart> >>> Bob La L>>>

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No, actually. He's solving a geometry problem, but he set up an algebraic equation.

It's only an approximate solution to the guy wire problem because a hanging wire forms a catenary not a straight line, so the actual wire will end up being somewhat longer.

Best regards, Spehro Pefhany

As a math problem it's inaccurate but as an engineering solution it tells you the minimum wire length. Then add 10% for the end fittings etc. Different mindset.

Uh, guys?

A^2 + B^2 = C^2 solves for the AREA of the Hypotenuse Squared...

Sqrt(A^2 + B^2) = C gives the length of the Hypotenuse. or, more familiar arrangement C= Sqrt(A^2 + B^2)

Richard