Riddles
- Thread starter Babak
- Start date
..Ok, I'll get to my point.
The amended problem CAN'T be solved with the lock starting dry since we don't have any idea as to the design of the boats. Are they flat bottom barges, deep v-hull speedboats, Boston Whalers, pickle-fork hydroplanes, washtubs or concrete canoes.
>If the lock had started at "empty" the displacement
>of a vessel is critical.
It's NOT the displacement that would become critical in this instance, but rather the design of both the vessel and the lock.
You stated:
>But speaking generically, it will take only 1/3rd of the
>water to float the 500t vessel as took to float the 15t
>vessel.
And by extrapolating those rough calculations it would take hundreds of times more water to float a tiny toy boat than it would to float the 15t boat, but we all know that this isn't true -- the toy model will float with just a thin layer of water on the bottom of the lock.
It's the shape of the boat and this shape relative to the shape of the lock that is critical to answering the amended riddle. In one case a higher displacement could take more water and in others less.
To further demonstrate this point using your water in the tub example, it truly DOES depend on the shape of the person where the water level will fall between the child and the adult. To give an extreme example, my Dad is a paraplegic whose legs and buttocks are literally skin and bones. If he drops in the tub the water may be LOWER than with the 30lb child and yet he weighs almost 200lbs. Also, if the child assumes a position that the adult is too large for (lies down for example) then that will effect the water level as well. Also, two people who weight the same will NOT necessarily make the water in the tub rise to the same level (women generally favor volume lower in the body than men).
Shape is everything.
JB
(your tub example and my responses really have little to do with the first problem since it doesn't involve "floating" anything.)
The amended problem CAN'T be solved with the lock starting dry since we don't have any idea as to the design of the boats. Are they flat bottom barges, deep v-hull speedboats, Boston Whalers, pickle-fork hydroplanes, washtubs or concrete canoes.
>If the lock had started at "empty" the displacement
>of a vessel is critical.
It's NOT the displacement that would become critical in this instance, but rather the design of both the vessel and the lock.
You stated:
>But speaking generically, it will take only 1/3rd of the
>water to float the 500t vessel as took to float the 15t
>vessel.
And by extrapolating those rough calculations it would take hundreds of times more water to float a tiny toy boat than it would to float the 15t boat, but we all know that this isn't true -- the toy model will float with just a thin layer of water on the bottom of the lock.
It's the shape of the boat and this shape relative to the shape of the lock that is critical to answering the amended riddle. In one case a higher displacement could take more water and in others less.
To further demonstrate this point using your water in the tub example, it truly DOES depend on the shape of the person where the water level will fall between the child and the adult. To give an extreme example, my Dad is a paraplegic whose legs and buttocks are literally skin and bones. If he drops in the tub the water may be LOWER than with the 30lb child and yet he weighs almost 200lbs. Also, if the child assumes a position that the adult is too large for (lies down for example) then that will effect the water level as well. Also, two people who weight the same will NOT necessarily make the water in the tub rise to the same level (women generally favor volume lower in the body than men).
Shape is everything.
JB
(your tub example and my responses really have little to do with the first problem since it doesn't involve "floating" anything.)
Ok, I'll give you an example of the opposite. In this case it will take MORE water to float the boat of the higher displacement.
The 15t vessel is a flat bottom barge that is the dimensions of the lock minus 1ft all the way around. (in other words...it JUST fits). It takes only a modest amount of fill before the barge diplaces 15t of water and lifts off the bottom and just "floats".
Now, load the same barge with 500t and it will sink and sit on on the bottom of the lock until MORE water is added to bring the displacement up to 500t.
JB
The chat has been fun, but for the record since the vessels are sitting still in the lock this issue has nothing to do with planing or displacement vessels. When both are sitting still there is only displacement at work and my responses only relate to that displacement.
Best wishes and thanks for the riddle.
JB
Best wishes and thanks for the riddle.
JB
Quote from sulong:
You're right.
I was working on the assumption of comparing displacement vessels to each other, and was not considering plaining vessels. (those are not a real sailors vessel anyway.)
Thanks for the chat,
Ok Gotcha sulong.Quote from sulong:
You're right.
I was working on the assumption of comparing displacement vessels to each other, and was not considering plaining vessels. (those are not a real sailors vessel anyway.)
Thanks for the chat,
I said Turok was right all along
still can't fathom that Fireboat not in water job though.
An now I got bloody Amoebas all over my keyboard and everywhere else too !!
Aphie, you should remember this! The old Bayes' Theorem :
P(A|B) = (P(B|A)*P(A))/P(B)
Let A = the event of having the disease
Let B = the event of testing positive
P(A|B) = (1*(1/500))/((1/500)+(499/500)*(0.05))
= 20/519
Notes
(1) P(A|B) is the probability that A occurs, given that B has occurred, known as conditional probability.
(2) Given no other information, I'm assuming there are no false negatives.
http://www.elitetrader.com/vb/showthread.php?s=&postid=177532#post177532
aphie and mr. subliminal are right
aphie said 4% which is approximately correct; mr sub came up with 20/519 which is even closer @ 3.85%
though i must say mr. sub you used a lot of high falutin' figgerin
easy way to figure it is like this:
if the screen test has a 5% false positive rate and 1 person out of 500 is actually infected, that means you will see 1 real case and 25 false positives in each hypothetical group of 500 tested
Therefore you can calculate the probability simply dividing 1 by 26 = 3.846
the key thing to remember is that the fixed percentage error rate shrinks or expands in terms of real world false IDs depending on the size of the control group necessary to generate one real case.
if the true statistical occurrence of the disease is low enough, the error rate of false positives could still dwarf the number of actual cases even if the test is super accurate.
so you could turn up positive on a test w/ 99.5% "accuracy" (0.5% false positive rate) and still have a very strong chance of not having the disease...
p.s. for the sake of simplicity false negatives weren't included but in the real world they only serve to make the numbers even muddier... lies damn lies and statistics :eek:
aphie said 4% which is approximately correct; mr sub came up with 20/519 which is even closer @ 3.85%
though i must say mr. sub you used a lot of high falutin' figgerin
easy way to figure it is like this:
if the screen test has a 5% false positive rate and 1 person out of 500 is actually infected, that means you will see 1 real case and 25 false positives in each hypothetical group of 500 tested
Therefore you can calculate the probability simply dividing 1 by 26 = 3.846
the key thing to remember is that the fixed percentage error rate shrinks or expands in terms of real world false IDs depending on the size of the control group necessary to generate one real case.
if the true statistical occurrence of the disease is low enough, the error rate of false positives could still dwarf the number of actual cases even if the test is super accurate.
so you could turn up positive on a test w/ 99.5% "accuracy" (0.5% false positive rate) and still have a very strong chance of not having the disease...
p.s. for the sake of simplicity false negatives weren't included but in the real world they only serve to make the numbers even muddier... lies damn lies and statistics :eek:
>if the screen test has a 5% false positive rate and 1
>person out of 500 is actually infected, that means you
>will see 1 real case and 25 false positives in each
>hypothetical group of 500 tested
>Therefore you can calculate the probability simply dividing
>1 by 26 = 3.846
Sure enough, and it even make sense to this dimwit.
JB
>person out of 500 is actually infected, that means you
>will see 1 real case and 25 false positives in each
>hypothetical group of 500 tested
>Therefore you can calculate the probability simply dividing
>1 by 26 = 3.846
Sure enough, and it even make sense to this dimwit.
JB
Quote from darkhorse:
aphie and mr. subliminal are right
aphie said 4% which is approximately correct; mr sub came up with 20/519 which is even closer @ 3.85%
though i must say mr. sub you used a lot of high falutin' figgerin
easy way to figure it is like this:
if the screen test has a 5% false positive rate and 1 person out of 500 is actually infected, that means you will see 1 real case and 25 false positives in each hypothetical group of 500 tested
Therefore you can calculate the probability simply dividing 1 by 26 = 3.846
Hmmm, Mr. Subliminal was more correct than me in this case, because I was thinking along the lines of Bayesian statistics but my interpretation of it was wrong.
I figured the following:
We know for a fact that out of a sample of 500 people, 1 will definately be infected.
We also know that out of 500 people, the false positive will show that 25 people have the disease.
Now, where I made the error was in taking 1/25, but I forgot to add the extra case where someone is ACTUALLY infected. Oops.
1/25 vs. 1/26
25 false positives, but there will be 1 correct positive per 500 -- for a total of 26 positives per 500 tests. My bad.