Obviously lots of “family” men and “card-sharps” in this thread, but no window cleaners or “real” construction engineers with their ladders (apologies Mike !!)
Using the starting formula below for L^2, and a scaled drawing to get a realistic starting value for “d”, it should be relatively easy to get a 3 sig fig answer for a 5m ladder. Even starting with 0.1; 0.2; 0.3 etc should soon get things underway.
The above solution is, in my opinion, the most easy to derive and the most easy to comprehend.
It does require a neat little substitution in the equation that I published a few posts above. But at least you all now know one destination. As a check, for a 10m ladder, d = 0.1119m to 4 sig.figs
There are other formulae, but they all generate the same output. If you superimpose the graphs in Excel, you can’t distinguish one from the others.
Some people seem to think that this later solution is “neater” than the previous one. I don’t.
Anyhow, you now have three possible mathematical formulae for solving for “d”.
Try working out a few values of “d” for increasing ladder lengths from say 3m up to say 10m. Note, the shortest ladder possible is 2.82m (ie 2*root2). Use each formula. The position plots from all three formulae should all lie on the same curve.
Re:- Ladder Problem
Unlike many of Don’s puzzles, which often have an elegant or simple solution, the solution to the ladder problem seemed to be much more complicated than one might have initially thought. I remember thinking the same when it was first posed some years ago, but I was no nearer finding the answer this time round either.
Thanks Don for keeping our brains ticking over.
Glad you remembered Brian’s (BAM) Brain Teaser from all those years ago. It took us
quite a while to figure it out back then, but eventually got the more straightforward solution (the one without the Acosh !). ISTR that Bam wasn’t overly impressed with that solution and was looking for something more “elegant”.
It was Ken C (IIRC) who came up with a solution incorporating Acosh, but by them BAM had disapperaed !! So we never really found out what sort of “elegant” solution he had in mind.
I’m not sure if my posted Acosh solution is the same as Kenc’s but it’s the only version that I can derive at the moment. The more straightforward solution that I posted is easy for me to remember, you don’t forget those struggles very easily !
Glad to know that this thread keeps a few grey cells nicely oiled !!
I was walking along a footpath at three and a half miles per hour the other day when a car whizzed past me and missed me by only a few inches.
“What speed was he doing ?” asked Mrs D who was beside me but away from the car.
“Well” said I, (twenty seven steps later), as I noticed the car had just disappeared around a corner, “I’ll tell you in a moment.”
We reached that corner another one hundred and thirty five steps later. I said to Mrs D, “Assuming that both we and that car were each travelling steadily at our respective speeds, and assuming that our paces have been uniform throughout, that car must have been doing …………”
If the couple had walked the same 27 steps before the car disappeared around the corner, but then had to walk twice as far, ie 270 steps to reach the corner, how fast would that car have been travelling ?
Same rules apply, ie all steps are uniform and all speeds are steady.
She said that on 1st January, the day after her birthday on 31st December.
So assuming this happened this year:
30th Dec ‘19 she was 13
She turned 14 on 31st Dec ‘19, and made the statement 1st January 2020. She turns 16 on 31st Dec 2021, “next year”.