Brain Teaser No 1

if you multiply 0.33333... (which is EXACTLY the same thing as 1/3) you will get 1, which is EXACTLY the same thing as 0.99999.... the recurring bit is important.
this comes from the sum of an infinite GP whose first term is 0.9 and common ratio 0.1, as i believe you worked out in your last posting.

try to start your multiplication of 0.33333.... at the LAST decimal point!!! its easy to go to infinity through taking limits, i am not sure what mathematical device you would use to start from infinity and work backwards. maybe there is one, i dont know it...

enjoy

ken

Rsults to be posted...

Matthew

1/17 must just be a coincident.

Oh well, back to the ladder!

Matthew

a lorry is travelling in a straight line at constant speed. it starts turning so that the front wheels describe a circle of some radius.

question: is the imaginary line along which the lorry was travelling originally a tangent to this circle?

if so, why so, if not, why not...

i am quite sobber right now, so i should be able to get my head round whatever you send...

enjoy

ken

This gets all rather complicated. each wheel will discribe a different circle, inside front will discribe a smaller circle then outside front.

Anyway the, the answer I guess is no. The truck when it starts to turn will not discribe a circle but some other strange shape which I don't think I feel inclined to figure out yet.

cheers

Matthew

quote:

---

Originally posted by Matthew T:
ken c

This gets all rather complicated. each wheel will discribe a different circle, inside front will discribe a smaller circle then outside front.

Anyway the, the answer I guess is no. The truck when it starts to turn will not discribe a circle but some other strange shape which I don't think I feel inclined to figure out yet.

cheers

Matthew

---

no wonder there was quite a heated argument in the pub when this question was raised. as i said, i was well gone by then and i recall wondering what the fuss was about. at the time i thought the answer was obviously yes, but i dont have any model to base this on... if you come up with one, let me know, but i wouldnt plan your life round this one...

many thanks matthew...

enjoy

ken

So I think the answer is YES for any single point on the lorry. Imagine you have a toy lorry sitting on a large, horizontal whiteboard. Using a pen you draw a circle and a tangent. Mark a point on the toy lorry. Now is it possible to navigate the lorry so that the point is always directly above the line and circle?

BAM

Under normal circumstances, lorries, trains aeroplanes etc do not make a sudden transition from a straight line to a pure circle or vice-versa. The straight line and the pure circle are linked by a "transition curve". In the case of trains, it is the track that incorporates the transition and the train just follows the track.

The IDEAL transition is an Euler Spiral (theta=d^2/2RL) where theta is deflection angle, d is distance along transition, R is radius of pure curve and L is length of transition. This is begining to remind me of that bl**dy ladder!!.

The transition is designed to be tangential to both the straight and the curve.

No matter how quickly the lorry turns its wheels, there will be a transition. It might not be very long and it almost certainly won't be an Euler Spiral, but it will be a transition curve!

The effect of a transition is to "shift" the pure curve to one side, so the straight is not tangential to the pure curve.

Now I think I'll pop down to the pub for a pint.....or on second thoughts, maybe not, just in case ken c's mates are down there!!!!

cheers

Don

But it seems to me that if we attach a line segment to a curve with finite rates of change then the line is tangent to the curve. And a truck cannot change direction with an infinite rate of change.....

Paul

a few posts ago you said i sort of suspected this curve was a hyperbola, but i am not sure where that leads me...

Well, i've tried plotting this curve (given up on formulae after the Euler Spiral) and it doesn't look like a hyperbola to me. It almost looks like a semi-circle, but it clearly isn't and it also closely resembles a parabola, but probably isn't.

I plotted the curve starting with a vertical ladder and stopped just as the top of the ladder was about to fall off the crate!

I can claim to be sadder, but certainly not wiser!

OTOH, I noticed that if you were standing ON the ladder and the toe of the ladder slipped away from the wall, and the top slid down the wall, then YOU would describe a parabola. This suggests that PROVIDED you hang onto a sliding ladder, you should have a fairly soft landing. Any volunteers???(Bam?)

Cheers

Don

I'm playing about with the numbers, and getting pretty graphs, but still unable to find an elusive formula.

Could be. I've tried 'e' and 'pi' in various parts of different formulae, all without success. Bam claims his solution is elegant. 'e' and 'pi' are sort of elegant, but the Sony engineers would be bu**ered without their log tables or a calculator if 'e' or 'pi' featured. At least with our 'crude' formula, you can get an answer in less than five minutes for any L. No doubt Bam's formula gets the answer in 30 seconds without needing tables or square roots.

Bam?

Cheers

Don

As anyone can see from the attached foumula, the obvious answer is evident,

Nice one! I think you managed to get us all certified at once with that!!

Mind you, at 81kg I now know which wave-length I'm on, but I haven't got a clue what wave-length Bam, ken c, Paul R, Omer, Steved, Mattew T et al are on!

Slowly going round the bend

Cheers

Don

Quote from a couple of pages back...I said to ken c...

Mind you, I have also plotted the curve using the 'precise' formula that you published. If L is plotted on the X-axis and d is plotted on the Y-axis, then L is asymtotic to the X-axis at y=0 BUT d ISN'T asymtotic to the Y-axis at X=sqrt(8), because at this point the tanget to the curve is parallel to the Y-axis.

The curve looks like a hockey stick.

Of course, I plotted it on normal squared paper, not log or log/log paper...silly me!

I had plotted this two months ago, before Xmas (just how sad have we all become?) by using L^2=(1+d)^2 + (1+1/d)^2. Both curves look the same.

Still can't see the light

Principal values that I used are

L=2.828;2.9;3.0;3.1;3.2;3.3.3.4;3.5;4;4.5;5,6,7,8,9,10,15,20,30,40,50,100

d=1.0;.7717;.6702;.6053;.5569;.5181;.4859;.4584;.3622;.3024;.2605;.2050;.1694;.1446;.1261;.1119;.0716;.0527;.0345;.0256;.0204;.0101

Any help?

Cheers

Don

Saved you a pint and Fermat lives!!!

BTW, if you were at the 3.30 Forum session, I was the prat at the front right who asked about phono boards in the 552 (well I felt sorry that there weren't too many questions being asked!) and was told (by that kindly gentleman from Naim, Paul D, who must wonder sometimes just how banal our questions could possibly get), there weren't any, you need a Prefix. I wonder if my Linto would do to start with? hee hee.

Cheers

Don

Even tempted to go and by an a-level further maths book to see if I might be right, probably would be sadly disappointed!

cheers

Matthew

PS if all those wheels carried on on a tangent to there curves the lorry would not remain the same size

Now, i haven't thought the next variant through at all (believe me!) but if we put five (5) circles inside the big one, do you recon there's a "golden ratio" in there somewhere?

Well, I've now thought about it and GUESS WHAT!!

Yep, you're all dead right, there is a beautiful golden ratio sitting in there just waiting to be discovered.

For those who are asking " what's a golden ratio, remember 1.618... and 0.618....??"

Have fun!!

Cheers

Don

Yep, you're all dead right, there is a beautiful golden ratio sitting in there just waiting to be discovered.

wow!!

i will have a look again via paul's general formula.

enjoy

ken

enjoy

ken

Where R is the radius of the encompassing circle, r is the radius of the internally disposed circles, x is the radius of the small centered circle and A is 180/n degrees, where 'n' is the count of small circles,

r = RsinA/(1 + sinA)
x = R - 2r

I noticed that x can also be expressed as

x = R(1 - SinA)/(1 + SinA)

I wonder if Bam (or Paul) think this is more elegant ??

Cheers

Don

Anybody see any 'golden ratios' in the five circle version of ken c's wheels within wheels?

Remember the golden ratio is 0.618... and 1.618...which are derived from

0.5[1 +/- sqrt(5)] or

0.5[sqrt(5) +/- 1]

Cheers

Don

Parabolic belly flop/face plant

I've been trying to follow all these graphical pursuits and I think this approach is laudable. It isn't the approach I had in mind so I am happy to comment on it. It has potential to lead to a simpler solution - if you can identify the function(s) that fits the graph.

I realised that my 'hocky stick' curve d v L, for which I posted a dozen or so points, could be turned into a nice looking parabolic type of curve, by plotting a mirror image above the y = 1 line.

Horizontal parabolas, shining to the right and centred on y = 0 (axis) have a general formula
y^2 = 4ax (a = focal length)

We would then be looking for the equation of a parabola in which the vertex is at (sqrt8,1), and with y = 1 as the horizontal axis and in which 0 < y < 2. We would only be interested in the lower half of the parabola where y < 1.

OTOH, the curve might just be more like a hyperbola!!! or any old quadratic equation!!!

Wonder what sort of curve Bam would have followed from the 10th floor Presidential Suite of the Bristol Meridian.........

Cheers

Don

quote:

---

x = R(1 - SinA)/(1 + SinA)

I wonder if Bam (or Paul) think this is more elegant ??

---

Errm. I'd obviously decided to leave the tidying up as an exercise for the reader....

There's probably something profound to be learned from considering that equation as a continuous function.

Paul