Workshop garrotte

[mklotz]

Marv

Due to the late hour and heavy eyes, I misstated my solution.  It should have been: ...

Proofreading is a virtue.

Your equation is a quartic.  Are you going to show us how you go about solving for A?

[Captain Jerry]

Marv

You asked for the reasoning, not the solution.

I'm going to go strap some wings on the donkey.  I don't think it will fly, but if it does, I will be able to show the solution.

Jerry

[mklotz]

Re:- The maths , maybe I have mis-read this ,maybe I don't understand American, but I'm lost. Used to think I was pretty good maths. But keep 'em coming, good to think now and then.

Just be glad that I didn't request that all the arithmetic be done with Egyptian fractions.

[mklotz]

Marv

You asked for the reasoning, not the solution.

I'm going to go strap some wings on the donkey.  I don't think it will fly, but if it does, I will be able to show the solution.

Jerry

If we convert your equation to its canonical quartic form, we have:

A^4 + 12*A^3 - 553*A^2 + 432*A + 1296 = 0

Plugging these coefficients into my 234 program, we get the following solutions...


Order of equation to solve (2,3,4) [2] ? 4

For order 4 term, input coefficient

• ? 1

For order 3 term, input coefficient

• ? 12

For order 2 term, input coefficient

• ? -553

For order 1 term, input coefficient

• ? 432

For order 0 term, input coefficient

• ? 1296

Solutions to:
 +1.0000 * x^4 +12.0000 * x^3 -553.0000 * x^2 +432.0000 * x^1 +1296.0000 = 0
are:
real:  17.672904  imaginary:  0.000000
real:  2.037017  imaginary:  0.000000
real:  -1.179138  imaginary:  0.000000
real:  -30.530782  imaginary:  0.000000


Note that the first solution (17.673) is the correct answer when added to the box height.  The second solution, when added to the box edge, is the (correct) distance of the bottom of the ladder from the wall.  [This is the other solution when the problem is rotated 90 degrees.]  The last two solutions are extraneous and can be disregarded.

Well done, Jerry.  In fact, it's your approach that I used when writing the program to solve this problem for arbitrary box and ladder sizes.

[Captain Jerry]

Marv

You might not believe this but the donkey took to the wings like a natural athlete.  After only two practice attempts, he gained enough altitude to clear the fence. Unfortunately, he did not have enough altitude to clear the tree line and wound up in the mid-level of an ancient live oak.  It took me most of the afternoon to talk him down.

On the way back to the house, I remembered my promise to walk you through the solution.  I guess I took too long playing with the donkey, because when I logged on to post the solution, I saw that you had already figured it out for yourself!  Nice going!

Jerry

[tel]

Well y'awl lost me a couple of pages ago BUT .... as far as I can see the problem has not been resolved at all. Take a look at the question again;

A 6 unit cubical box is pushed tight against a vertical wall.  A 25 unit long ladder is placed against the wall such that it just touches the edge of the box.  What is the height of the point at which the ladder touches the wall?

Nowhere is the height of the wall specified! Now if the height of the wall is greater than then length of the ladder, all well and good BUT if the wall happens to be shorter by any significant amount then H=h. No variation is possible.

[mklotz]

Thanks so much for that, Tel.  I just don't know what we would do without these penetrating insights of yours.  Well, actually, I do know but I'm keeping it a secret lest it elicit another penetrating insight.

[tel]

Just thought you should know - there was a risk that you might try to lean the ladder on the wall at a point avove the top, and someone might get seriously injured! 

[Mosey]

Tel,
Try to avoid complicating the logic with facts.
Mosey

[Maryak]

OK,

I used the cheats way

 

Sin B =	6/Z	Cos B =	Y/Z
Sin B =	X/A	Cos B =	6/A
Sin B =	(X+6)/25	Cos B =	(Y+6)/25
	Assign to B	71.2475	deg
0.9469	= 6/Z	0.3215	=Y/Z
Z =	0.1578	Y =	2.0370
	X+6 =	23.6729
	X =	17.6729
Tan B =	2.9455
B =	71.2475
	Solved by Iteration!!

Best Regards
Bob

[Mosey]

It can be solved rather quickly graphically (the Hardy Cross method), though obviously not exact.
Mosey

[mklotz]

It can be solved rather quickly graphically (the Hardy Cross method), though obviously not exact.
Mosey

And you'll learn nothing about mathematics by doing so.  Remember this from the original statement of the problem?

Remember, the numerical answer is not the object here.  Any fool with a CAD program can find that.  What is required is the mathematical reasoning that leads to the answer.  The numerical answer is then just a quick way to check the validity of the reasoning.