The Crossing Ladders

Hello ba/ael,

I see there has been a few readers, but no replies, so this is the first.

Specification Problems, once again.

Following your detailed Specification as above, I cannot see how the ladders can intersect, because they effectively become useless except to monkeys, ants, ivy, and the like, as I am now not such a good climber..

If they do intersect, the stiles of each ladder actually pass through the other ladder's stiles.

Thus they could no longer be correctly described as ladders.

Are the walls exactly parallel in all planes, or are they vertical (remember the spherical Planet - there is a difference), or are they just a line on graph paper, which I cannot properly see in my monitor?

Do you intend just a geometry/trigonometry puzzle, on a piece of paper?

Advise please.......

Sparky,

The ladders, as you have so astutely noted, do not really intersect, but in elevation, they appear to intersect. If you prefer, call them string lines. Even they don't intersect as they are offset by the diameter of the string.

The walls are vertical. It is not a trick problem. In fact, it is a straightforward geometry problem which anyone can solve by iteration. I don't know how to come up with a closed form solution for the distance between walls. I was hoping that the boys with MatLab might be able to humour me in this regard.

Have fun! If you dare!

Sparky, I posted this problem months ago. It has been discussed in great detail here.

-John

This is similar, but not the same problem. The currrent problem is MUCH harder for reasons discussed in repllies #25 & #26.

What is the horizontal distance between the two walls?

Wherever the ceiling is.

cr3

this is a really simple trig problem unless i"m missing something. Do I have to go get my calculator?
Or can I cherry pick?

It may not be as easy as you think, CR3. If you call the distance between the walls 'x' and the ladder lengths 'a' and 'b' and the height of intersection 'h', can you come up with a solution for x in terms of a, b and h?

If you can, then I would like to see it, because I can't seem to find it.

I started to work long hand and quickly saw I was a known short......

I'll get back with you on this one...dowels? no. cad? no. Perhaps some pasta? Hmmmm...

I just cut two pieces of dowel 20" and 30" thus forming a rudimentary analog computer...I bet I got my answer quicker than you guys (yes I did remember to scale the answer back up) .

(May not have quite the accuracy, but was damn quick.... by the way..anyone want a couple of short bits of dowel?)

12 ft - 3 inch, using cad

Regards JD.

Close enough, jd. But I'm waiting to see someone come up with an analytical solution other than successive approximations, which is okay as an engineering technique, but doesn't quite have the pizazz of a unique solution. I must tell you in advance that I don't know the answer.

Hello ba/ael,

I have spent some time on this, and have a proper analytical solution other than successive approximations.....

Hehe hehe......

Kind Regards....

Hello Sparkstation,

To get you over your ennui, why don't you spend a little time sprucing up your windmill? Maybe a coat of paint or something.

The attached is just a suggestion. I'll leave the animation up to you.

Hello again Sparkstation,

On the other hand, you may prefer one of our more modern models (see attached). They operate best in the presence of a lot of wind. In your case, there should be no problem.

Best regards,

Hello again ba/ael,

Sorry I didn't reply sooner.

There was a lot of wind here today, and I have been rather busy, got some cheap left-over paint from a friend.

I followed your wonderful advice, and started to paint that first windwill you sent me.

Not all the paint stayed on the windmill, apparently because it was slow-drying enamel, left over from some international exhibition of modern art.

Once again, I have gotten myself into trouble, both with my long-suffering wife, and the neighbours, who are normally a passive lot.

I tried to avoid their rude remarks soon after I started today's efforts, because I realize the fickleness of others, who often change their opinions as fast as the wind direction around these parts.

I cannot see what the trouble is, after all the windwill, now fully repainted, looks like new.

As the increasing wind rotated the blades just after I completed painting the last blade, the windmill brake unfortunately released at the same time, and it seems that the uncured paint flew off in large gloppy droplets, and rained down on the bystanders and their houses etc.

I got tarred and feathered and run out of town on a rail.

Unfortunately I dropped my paintbrush during my travels, and ask you to pick it up, and place into some turpentine, so the bristles don't harden with the high-gloss enamel paint I so carefully applied.

There it is, on the ground, just below.

Kind Regards....

.

Hi, Sparksperson,

You don't have to apologize for not responding sooner. We all know that people of your vintage require lots of sleep. And, to assist in this regard I am forwarding this idyllic little scene of sheep jumping over a fence.

I have never found that it works for me, but I don't live in sheep country. I am told that you do. If you focus carefully on the sheep in the attached picture and try as best you can to count the little fellows, you will have a sleep to remember.

But, enough about that. Let me apologize for my advice about painting your windmill. I was under the impression that you were in the contracting business and would be familiar with the proper paint to use. I particularly apologize for the neighbours rude remarks. I didn't think New Zealanders carried on that way.

I see your paintbrush in the picture, but I can't seem to find it on the ground. I will keep looking and let you know when I find it. In the meantime, keep smiling!

Best personal regards,

Hello Sparky,

I found this picture of a ladder which I felt I should forward to you when I learned of your, um...painting difficulties. The idea is to ascend to the top of the ladder, one step at a time starting with the rung nearest the bottom. If you have a problem carrying the brush and paint can while climbing the ladder, you may place the can and brush on the ground near the ladder with a rope attached thereto, ascend the ladder while grasping the rope, then, when you get to the top, hoist both can and brush to the item to be painted while taking great care to maintain your balance.

Please log on to our website and see our video illustrating the recommended method of raising the paint can and brush. Today, with all the litigation that goes on, we simply cannot be too careful, especially with our beloved seniors such as yourself.

I sincerely hope that this will assist you in your next painting endeavour.

All the best to you,

PS I still haven't found your paintbrush, but don't worry...the kids in the neighbourhood have promised to help. If we can't find it, they have promised to buy you a new one.

Is this a homework question?

I gave it a try, but damn i´m rusty. Anyhow...

Nomenclature

AB= line1

CD= line2

a= CAB angle , b= ACD angle

d= distance AC

I = the distance where the vertical of the E point intersects the horizontal (square triangle).

Equations based on trigonometry:

d = 20 cos(a) eq1

d= 30 cos(b) eq2

d-I = cotg(b) eq3

I = cotg(a) eq4

Substituting I of eq4 in eq3:

d= 10cotg(b) + 10 cotg(a)

we get

d= 10 {cotg(a)+cotg(b)} eq3a

applying the following trigonometric identity...

cotg(a)+cotg(b) = sin ( a+ b) / {sin (a) * sin (b)}

we get:

d= 10 * sin ( a+b)eq3b

sin (a) * sin (b)

so, together with

d = 20 cos(a) eq1

d= 30 cos(b) eq2

we have a 3X3 system. Is it solvable by numerical means?

specially because:

cos (b) = 2/3 cos(a)

Then we get a 2X2 system:

10 sin (a + b)/(sen(a)*sen(b)) - 20 cos(a) = 0

10 sin (a + b)/(sen(a)*sen(b)) - 30 cos(b) = 0

So for any dimensions we get the following equations, ready to graph:

In this example, segment EI = 10.

segment EI sin (a + b)/(sin(a)*sin(b)) - length of AB * cos(a) = 0

segment EI sin (a + b)/(sin(a)*sin(b)) - length of CD * cos(b) = 0

Which can be "simplified" like this:

equalling eq1 and eq3b

d = AB cos(a) = EI sin(a + b)/(sin(a)*sin(b))

AB / EI = sin (a + b)/(sin(a)*cos(a)*sin(b))

Trigonometric identity --> sin(a)*cos(a)= sin(2*a)/2

so we get

AB / EI = (1/2) sin(a + b)/(sin(2*a)*sin(b))

and finally:

sin(a + b) / (sin(2*a)*sin(b)) - (1/2)AB/EI = 0 (A)

----

and simetrically, doing the same, equalling eq2 and eq3b:

d = CD cos(b) = EI sin(a + b)/(sin(a)*sin(b))

we get:

sin(a + b) / (sin(a)*sin(2*b)) - (1/2)CD/EI = 0 (B)

:P

Well Stated!

Two comments:

1. Once you have solved for a, then you can solve for x directly by using equation 4, solving for b is unnecessary.

2. You were vague about the solution of the quartic polynomial equation for a. Iterative methods work, of course, but there is an ancient formula for solving quartics in closed form: http://mathworld.wolfram.com/QuarticEquation.html

If you don't want to do the algebra, then Mathematica will provide the closed form solution.

Hi fwes,

Thanks for the comments.

In 1, I purposely solved for b (b=27.35723252') to show the validity of the value being, 10 < b < 30, when the value of a (a=15.76128711') being 10 < a < 20.

In 2, I solved the quartic equation for a values using the function "Polynomial Root Finder" with my TI-83 Plus scientific calculator. The solution is direct and the a values are not results of approximation.

You may find this similar thread of interest.

You might also consider buying a tape measure.

Ken,

You are right. The other thread is similar, but that problem is easier to solve than this one.

Suppose we use 'x' to represent the distance between walls. The ladders are of length 'a' and 'b' and they intersect at height 'h'.

Let h1 and h2 be the upper elevation of each ladder.

We can say that: (1) h1.h2/(h1 + h2) = h (similar triangles)

Also, (2) h1 = (a² - x²)^1/2 and (3) h2 = (b² - x²)^1/2 (Pythagoras)

If a, b and x are given, h1 and h2 can be solved directly and, from that, h.

However, given a, b and h, it's not so easy to separate variable 'x'. Equation (1) can be inverted to give (1a) 1/h1 + 1/h2 = 1/h, or (1b) h/h1 + h/h2 = 1 but it doesn't remove the problem.

The other thread is similar, but that problem is easier to solve than this one.

True.. although in post 21 of the other thread, Andy H snuck in your problem... which Willyap06 quickly solved in post 23, (although without showing his work).

These problems are fun -- they seem so simple at first glance, and then you start to say "Oh... wait... something's missing." In the real world, I often do these things (i.e. like yours -- not the other thread) iteratively... but I feel sort of guilty, and am glad I don't have one of my math teachers looking over my shoulder. Excel has fairly well destroyed any competence I once had in math.

These problems are fun -- they seem so simple at first glance, and then you start to say "Oh... wait... something's missing." In the real world, I often do these things (i.e. like yours -- not the other thread) iteratively...

Uhhh, guilty as charged sir. Now and again I am afraid.

Good work, guys! I have given two "good answers", one to post #20 by willyap06, the other to post #23 by Doppel.

Nearly the same thread by IronDoubleEE: Unequal Ladders in Alleyway

Top marks, Abdel Halim Galala

It is true: "There is nothing new, under the Sun".

Kind Regards....