Random Questions.....

[peter]

......not worth a thread of their own thread.

Any body know how many [about] words they read a minute [when reading in a relaxed manner from say, a novel you are reading at bed-time or whatever?

{Mods - if we have a 'random questions thread already (I suspect we may but couldn't find it), by all means dump this question into it.}

[Avatar]

About 1,200 words per minute. Or an average of about 120 pages an hour. I usually test every few years to see if it's changed, but it's been steady for well over 10 years now.

--A

[peter]

Woah! By that standard I can barely read at all!

[Avatar]

I get a lot of practice.

--A

[peter]

Seems like the answer to this should be common sense, but will a given area of cardboard, if made into boxes of different dimensions [by making rectangular boxes as well as the single square one that could be made] always make boxes of the same volume [eg 6 meters sqare -> 1 meter cubed]. If not would the smallest volume alwas be the least tall box with the largest surface area top and bottom. [Is the Calculus required to answer problems where more than two variables need solving, which I think might be the case here?]

[wayfriend]

Nope. Given the same surface area, the closer the box is to equal dimensions, the more volume it contains. In two dimensions, given the same perimeter, the closer the rectangle is to equal dimensions, the more area it contains.

[Hashi Lebwohl]

Wayfriend is correct. It is relatively easy to show for 2-dimensional objects.

perimeter = 2(length + width) and is a fixed number; area = length * width

a = l*w but p = 2(l + w) --> w = (p - 2l)/2
a = l * w --> a = l * (p - 2l)/2 = (pl - 2l^2)/2 = (p/2)l - l^2
now differentiate
da = (p/2) - 2l and to find maximum volume presume da = 0
0 = (p/2) - 2l --> l = p/4
p = 2l + 2w = 2(p/4) + 2w --> p = p/2 + 2w --> p/2 = 2w --> p/4 = w --> l = w --> for a given perimeter squares give maximum area

It is trickier to show for 3-dimensional objects

A = 2lw + 2wh + 2lh (l = length, w = width, h = height, obviously) and V = lwh

the process is the same, though--A is fixed so we can solve for either lw or wh or lh then sub that in to V, taking us from three variables to only two, then differentiate and set dV = 0. We will eventually arrive at the condition that l = w = h for a cube giving maximum volume for given surface area A.

Do I really have to work out those steps? I used to be a Level 7 contributor in the mathematics section of Yahoo Answers but that was quite some time ago.

edit: I figured out a decent way to reduce the complexity of the problem. We already know that for a given perimeter a square gives maximum volume, so let us presume that the end of the rectangular solid is a square: h = w (we will refer to their value as k)

A = 2hl + 2hl + 2wl = 2k^2 + 4kl --> l = (A - 2k^2)/4k
V = lwh = lk^2 = k^2 * (A - 2k^2)/4k = (Ak - 2k^3)/4 = (A/4)k - (1/2)k^3
dV = (A/4) - (3/2)k^2 and again maximum volume at dV = 0
(3/2)k^2 = (A/4) --> A = 6k^2
l = (6k^2 - 2k^2)/4k --> L = k
so now we find that when the end of the solid is a square maximum volume is attained when all three dimensions are the same --> a cube

We would still have to solve the more general case when w <> h but that got me to

w = -(l+h)/2 + sqrt[A + (l+h)^2]/2

[Avatar]

WTF?

--A

[peter]

Wow! I didn't see that coming Hashi! Is that calculus? Also, that relationship of the volume of a cube being always one sixth of it's surface area - but raised a power. Does that hold the for all regular sided solids (but with different values obviously). What say would it be for a triangular solid composed of four equilateral triangles (I think such a solid exists)?

[Hashi Lebwohl]

WTF?

--A

I often have that effect on people.

Wow! I didn't see that coming Hashi! Is that calculus? Also, that relationship of the volume of a cube being always one sixth of it's surface area - but raised a power. Does that hold the for all regular sided solids (but with different values obviously). What say would it be for a triangular solid composed of four equilateral triangles (I think such a solid exists)?

The differentiation step is calculus; the rest is just algebra.

A mathematical solid composed of four equilateral triangles is a regular tetrahedron. From that page you can get to links for the other regular solids to look at the relationship between surface area and volume. There really isn't anything special about any of them except for the sphere--for a sphere if the take the derivative of the formula for volume (4/3)πr^3 you get 4πr^2, which is the formula for surface area of a sphere.

[JIkj fjds j]

On a clock, which hand is the right hand, and which is the left hand?


Edit: damn, that random synchronicity is uncanny!

[Hashi Lebwohl]

On a clock, which hand is the right hand, and which is the left hand?

There are at least two correct answers to this question: 1) whichever one you want to call the "right" hand and 2) neither one since they are technically "minute hand" and "hour hand".

Why don't we bother creating a new calendar that doesn't have alternating 30-day and 31-day months (February notwithstanding)? 13 months of 28 days each brings us to 364 then we just get a leap day every year and another one every 4 years--simple. Of course, who wants 13 Friday the 13th's every year?

[JIkj fjds j]

There are at least two correct answers to this question: 1) whichever one you want to call the "right" hand and 2) neither one since they are technically "minute hand" and "hour hand".

1) If we were to nominate the hour hand as right and the minute hand as left, wouldn't the hour hand only obey the rule for 6 hours (from 12 O'clock to 6 O'clock) to then become the left hand (from 6 O'clock to 12 O'clock). While the second hand, in vice verca, can only obey the rule for 30 minutes at a time.

therefore ...

3) Have the clock set next to a mirror so that both hands are always seen to be correctly on the right hand side and the left hand side, no matter what time of day they show the time to be.

[wayfriend]

On a clock, which hand is the right hand, and which is the left hand?

There are at least two correct answers to this question: 1) whichever one you want to call the "right" hand and 2) neither one since they are technically "minute hand" and "hour hand".

Pah!

As we all know, clocks are always pictured posing for 10 minutes to 2 o'clock.



So clearly the hour hand is the left hand, and the minute hand is the right hand.

[Sorus]

So the second hand is actually the third hand?

Oi, that's bad even for me.

Of course, who wants 13 Friday the 13th's every year?

No. Just no. That's my one and only superstition, but I take it seriously. Two out of three of the worst days of my life were on Friday the 13th.

[JIkj fjds j]

So the second hand is actually the third hand?

Yep, that makes perfect sense to me.

And as it's Halloween ... which is best for spells: digital watch, analogue watch, hour glass, or a sundial ?

[peter]

It stems from the day back in what? the 14th century when king Phillip the fair of France and the pope of the day rounded up the Knights Templar in order to crush the Order and seize their assets. Friday 13th was the date.

We live in the 'digital' age: will it always be the digital age from now on (in terms of tech and assuming we don't go back to the stone-age) or can there be a post-digtal advancement?

[Linna Heartbooger]

Seems like the answer to this should be common sense, but will a given area of cardboard, if made into boxes of different dimensions [by making rectangular boxes as well as the single square one that could be made] always make boxes of the same volume [eg 6 meters sqare -> 1 meter cubed]. If not would the smallest volume alwas be the least tall box with the largest surface area top and bottom. [Is the Calculus required to answer problems where more than two variables need solving, which I think might be the case here?]

Btw, the SAME DAY I saw this post, I did a similar problem with a student.

It was very similar, but the more traditional "if you START with a given rectangular piece of cardboard, and cut out 4 square corners, then fold the sides up to make an open-topped box," how do you get maximum volume?
(So they aren't all the same surface area, but there's a nice, straightforward connection between length, width, and height. Which means you can turn it into an expression in only 1 variable.)

So the second hand is actually the third hand?

I am frequently reminded of one time when Ananda was like, "On the third hand..." and always want to quote it cause it just seems like an expression that should exist. For hilarity's sake if nothing else.


Anyway, my wacky question is this: does anyone else here remember "The Internet Oracle," from back in the day?
(the canonical example of a Q to send the Internet Oracle being "How much wood would a woodchuck chuck if a wood chuck would chuck wood?")

[wayfriend]

It was very similar, but the more traditional "if you START with a given rectangular piece of cardboard, and cut out 4 square corners, then fold the sides up to make an open-topped box," how do you get maximum volume?

Yes, that's not quite the same constraint as having equal surface area. The smaller the corners you remove, the larger the surface area of the box. So the best result won't be exactly cubical --- but probably not far from. I hope those kids knew their differential equations.

[Sorus]

And as it's Halloween ... which is best for spells: digital watch, analogue watch, hour glass, or a sundial ?

I imagine the traditional spellcaster uses an hourglass or sundial, while the trendy modern spellcaster uses a digital device.

I've always preferred traditional methods. Say your battery goes dead right as you summon a demon lord from the depths of the twisting nether. That's going to be awkward, at best.

I am frequently reminded of one time when Ananda was like, "On the third hand..." and always want to quote it cause it just seems like an expression that should exist. For hilarity's sake if nothing else.

Ananda is awesome. I miss her.