Let's discuss Mathematics

[ParadigmShifter]

Square cylinders? What have you been smoking???

EDIT: Assume intake is unaffected by the geometry of the doob.

EDIT2: So let's say it burns linearly and all THC is inhaled.

 

[uppi]

Square cylinders? What have you been smoking???

Maybe he is using the uniform norm.

 

[ParadigmShifter]

Norm isn't here; he provided the dope and then left. He wasn't wearing a uniform.

 

[Tabster]

Aren't they both the same ?

volume of 1st half(cone-cylinder)= volume of 2nd half(cylinder-cone)

Have you been to see hawkwind again?

 

[ParadigmShifter]

I was careful to say they may be equal (who gets more stoned, and if anyone how much more stoned do they get?)

Not been to see Hawkwind for years

I haven't done the working but if anyone shows their working I will probably agree.

EDIT: I think you are right about the volumes being equal after drawing a diagram.

 

[ParadigmShifter]

So it looks like the answer is Ben gets more stoned earlier, unless he is smoking a pencil.

Looks like Abe only catches up at the last toke? (EDIT: Proof? Could be mistaken there)

 

[Tabster]

Ben - cylinder - gets more stoned

As the cone gets narrower the air flow through it isn't enough to allow full combustion, and its efficiency as a THC delivery machine decreases.

 

[ParadigmShifter]

Whoops got Ben and Abe mixed up in my previous post.

Airflow doesn't factor into it, we are assuming perfect smoking conditions and linear burn rate (m/s).

 

[Mise]

Oh yeah, cylinders are the ones with circular bottoms. Doh.

Anyway let r be radius at midpoint, l be length of spliff.

Volume of cylinder = pi*r^2*l
Volume of cone = 1/3*pi*(2r)^2*l = 4/3*pi*r^2*l = 4/3*Volume of cylinder

So whoever smoked the cone has a bigger spliff.

 

[Tabster]

hmm, I think your reasoning is right then

But more practical experiments need to be done in this particular field.

Edit : actually I'm not so sure, now.

@ Mise - 'square cylinders' - chessmate!!

 

[ParadigmShifter]

I think that is wrong Mise.

It's r rather than 2r for cylinder volume.

Also r = 2*l for midpoint.

 

[Atticus]

Mise had the cylindrer and ParadigmShifter the cone.

 

[ParadigmShifter]

What is the word Mise is looking for? (Like a cylinder, but a polygon base instead)?

Spoiler :

Prism

 

[Brighteye]

I am not sure what exactly you mean by the difference between them and I am assuming you mean the difference of the mean of both groups.

If so, I'd do the following:
1. Calculate the standard deviation of both groups.
2. Divide both by the square root of the respective sample size to get the standard deviation of the mean.
3. Multiply both with the Student's t-distribution (with a chosen confidence interval) to guess the standard deviation for an infinite sample size.
4. If you believe both groups to be independent from each other, add the resulting deviations quadratically (sqrt(x^2 + y^2))

That's how I would calculate the error in physics, but I am not sure it is done the same way in your field.

That seems good. Now I just need to work out what multiplying by Student's distribution is...

 

[uppi]

That seems good. Now I just need to work out what multiplying by Student's distribution is...

Student's t-distribution essentially takes two parameters: The sample size and the confidence interval. The latter is a measure of how safe the guess should be. Typical values are 1,2 or 3 standard deviations (68.3%, 95.4% or 99.7%). The sample size decreases the value of the t-distribution and a higher confidence interval increases the value.

Once you looked up the fitting values in a table or convinced you data analysis program to calculate it for you, you'll have a numeric value to multiply your standard deviation with.

 

[sanabas]

Bump for Paradigm Shifter...

From memory, this is called the hailstones problem, or series, or something.

Take any integer x

If x is even, halve it
If x is odd, then do 3x+1

rinse & repeat.

The conjecture is that it will always reach the loop 1-->4-->2-->1, but that's yet to be proven.

Prove that no matter which number you start from, you'll always reach a number divisible by 4.

 

[ParadigmShifter]

That is the Collatz Conjecture

http://en.wikipedia.org/wiki/Collatz_conjecture

 

[ParadigmShifter]

I've done one of the cases.

Assume that the initial number is not divisible by 4 (otherwise, it is immediately the case that you have a number divisible by 4).

So initial number is of the form 4k + m where k = 0, 1, ... and m is 1, 2 or 3

Case 4k+1:

This is odd. 3(4k+1) + 1 = 12k + 3 + 1 = 12k + 4, which is divisible by 4.

I'll look at the other 2 cases as well.

 

[sanabas]

Hadn't looked at it like that before, I suspect the third case won't be so neat.

4k + 2 --> 2k + 1, which must be one of the other two cases, so that can be ignored.


4k + 3 --> 12k + 10 --> 6k + 5. If k is even, then this becomes 4m + 1. But if k is odd, it becomes 4m + 3, and we're back where we started...

 

[ParadigmShifter]

Yeah I don't think it's going to be pretty for the 3rd case either after doing some scribbling