Brain Teaser No 1

a slightly better approximation for d:-

d=L-1-(1/L)+(1/(L^2)) (if your'e on the drunks ladder!)

and an even sightly better approximation for d:-

d=L-1-(1/L)+(1/(L^2))-(1/(L^4))

Cheers

Don

By longishI mean longer than about 5 meters

Below 5m the results are VERY crude.....but then the Sony engineers could probably get over the wall just using the cube.....so wouldn't need to know d.....ok, just a thought!!!

Cheers

Don

L=Length of ladder
d=distance from the cube

L^2=(1+d)^2 + (1+(1/d))^2 (Pythag)

L^2=d^2 + 2d + 2 + 2/d + 1/(d^2)

Which has a degree of 'symetry'

Also Let y = (d + 1/d) and hence y^2 = d^2 + 2 + (1/(d^2)) (1)

L^2=y^2 + 2y

0=y^2 + 2y - L^2 (2)

But this doesn't have simple roots!!!!!

Well the roots might not be simple BUT at least the problem CAN be solved by:-

Substitute for L in equ (2) and solve for y using the famous -b plus or minus sqrt......

Then substitute in equ (1) and solve for d using the famous -b plus or minus sqrt......

Not the most elegant solution, but at least the Sony engineers would sort their problem and find their roots.

BTW, there are normally 4 roots, two real and two imaginary, except when L=2*sqrt2 when there is only one real root (=1.000).

BAM, how much more elegant do you want the solutuion to be?????

Cheers

Don

Matthew

A planet which happens to have a small donut like tunnel within it which is perfectly centered on the center of the planet, which happens to be perfectly formed. For an object to maintain orbit in the tunnel what is the relationship between it's speed and the distance the tunnel is from the center of the planet.

The tunnel is small enough that the gravitional effect of it on the planet is nil, the planet is not spinning.

Matthew

Let me check my understanding (before I forget this and do some painting in my kitchen as a welcome alternative). The centre of the donut is coincident with the centre of the spherical planet. You want to know the relationship between the tangential speed of an object orbiting within the donut vs the radius of the orbit. You assume the volume of the donut is negligible. Is this correct?

Can we stay with Newton or do you require the full Keplerian proof? In this case, the answer is the same but do you want us to 'show our working'?

Best Regards,
Mark Dunn

Mark
I would be fascinated to see the full working, I was assuming a Newtonian approach.

cheers

Matthew

Help is needed. You both indicated that you had made some progress, publishing it would be helpful. It might be what Bam is looking for!!

My current thoughts are pretty simple, as outlined in previous posts

L^2 = d^2 + 2d + 2 + 2/d + 1/(d^2)

d = [y +or- (y^2 - 4)^(1/2)]/2

where y = [(L^2 + 1)^1/2] - 1

ooorrrr - in words

d = y + or - (sqrt of (y squared - 4)) all over 2
where y = (sqrt of (L squared +1)) - 1

Try writing it out - it’s a lot more simple than it looks!!!

Now in addition, it is obvious that Bam is looking for the "minus" sqrt of (y squared -4) solution

It is also obvious that the "plus" version and the "minus" version have to be reciprocals!!!! but I have yet to find a way of using this knowledge to simplify the solution. I can change it!! but make it more elegant??? no way!!!!

Any ideas????

I think I'll relax this evening with a decent record - how about Pink Floyd's "Another Brick in the Waaaaaaaaaahhhhh………..!!!"

Cheers

Don

Re:

>Mark
>I would be fascinated to see the full working, I >was assuming a Newtonian approach.

If I can find some time over the next couple of days I'll give it a shot. I haven't tangled with with curly ∂'s for a while but I'll see what develops.

Best Regards,
Mark Dunn@

quote:

---

Originally posted by Matthew T:
A planet which happens to have a small donut like tunnel within it which is perfectly centered on the center of the planet, which happens to be perfectly formed. For an object to maintain orbit in the tunnel what is the relationship between it's speed and the distance the tunnel is from the center of the planet.

---

Matthew,

I thought it was one of the standard 'givens' that an empty spherical shell had no gravitational effect on a body within the cavity. The mass of the part of the shell closest to you is exactly cancelled by the larger mass further away in the opposite direction.

That part of the planet's mass which is outside the orbit of the body can be considered such a shell, and thus ignored.

Thus, the problem is extremely simple. The solution is exactly the same as that for a satellite skimming across the surface of a body the same diameter as the orbital size.

cheers, Martin

[This message was edited by sceptic on SUNDAY 30 December 2001 at 15:08.]

For something to be in orbit its centripetal acceleration must equal its acceleration towards the centre of rotation. The centripetal acceleration is w^2*r (w=angular velocity and r=radius of orbit).

Let's say the acceleration due to gravity on the planets surface is g. Within the planet the acceleration towards the centre will be rg/R where R is the radius of the planet.

we have w^2.r = g*r/R
or w=SQRT(g/R) which is constant
So the tangential speed = r*w = r*SQRT(g/R)

So w^2*r = K^2*r^3 where K is some constant
w=K*r
tangential speed = K*r^2

What is the minimum number of weights needed to weigh up to 40kg in 1kg increments? Using the old 'Scales of Justice' or simple balance type of scales

I'll get back to the mind-bogglers later!

Cheers

Don

Mass of sphere = 4/3 x Pie x density x R^3
Gravitional acceleration = G x Mass of sphere / R^2
Centripedal accelaration = w^2 x r = v^2 / r

G is gravitioanl constant
R = radius of sphere
v = tangential velocity
w = angular velocity

Should be enough given the above postings, OK, trying to figure what Don is getting at!

cheers

Matthew

You COULD use 40 x 1kg weights but this wouldn't be the smallest number of weights.

You COULD use 7 X 5kg plus 5 X 1kg weights but....

Hope this clarifies

Cheers

Don

Now then, assuming we can put any of the weights on either side of the scales, the first two weights we need are easy. The next weight is then twice the sum of all the previous weights plus one kilo. This gives us a series of numbers and leads to the conclusion that we need 4 weights. Interestingly the series can also be represented by "powers of X".

Not being a mathmatical type, does this relationshpip of "powers of X" to the series "two times something plus one" occur often?

I'll leave X hidden for the moment so as not to spoil others' fun.

More like this please Don!

But does this method always give the least?

Paul

Re airplanes. I question whether they need cooling having sat frozen and squashed for hours on many an intercontinental flight. But if they do I guess I'd put it down to all the hot air generated by the passengers - and in my experience this is no trivial thing! And they recycle the air - yuk. I think fusealage heating through air friction is probably not so significant below supersonic speeds. A little heat may come from compressing outside air - but in my experience fresh air is not a significant feature of air travel.

Cheers

Keith.

Not being a mathmatical type, does this relationshpip of "powers of X" to the series "two times something plus one" occur often?

Never noticed it before, at least, not so as I remember noticing it! It only seems to be true for x = 3

If x = 2, then its only "the SUM of the previous powers plus 1"

And if x = 4, its "3x the sum of the previous powers plus 1"

And if x = 5, its 4x etc, etc

And, of course, if x = 1, its "0x the sum of the previous powers (which=0) plus 1". ie you go up 1 at a time.

I'm not sure if this helps us answer Paul R's subtle question - but it probably gives us a clue!

Any comments??

Cheers