Friday, December 16, 2011

How long is the movie

It is a tradition in our puzzle marathon to dedicate a special puzzle to anyone who earns 10 puzzle points. This puzzle is for and about our puzzle master, Jerome.

A friend of Jerome's presented him with a DVD of a special movie he burned for his birthday. It may be the never-seen goofy movie from Jerome's wedding or  last year's Holiday party where Jerome was Santa. Jerome is riding a subway home and wondering how long this movie is and whether he will squeeze watching it at home before going out to celebrate at the restaurant. On the front of the disk it is written that disk capacity is 120 min, 4.7GB.  On the back   there is a clear separation circle between the inner "written" part (lighter area) and outer "clean" part (darker area).  Jerome looks carefully at his DVD (pictured below) and sees a solution. What is it?

Answer ideas accepted any time until midnight on Sunday December 18th (EST), on our Family Puzzle Marathon. They will be hidden till then and everyone who submitted something reasonable will get a puzzle point.


anne-marie said...

We need to measure the surface of the entire outside circle.
We use the diameter of the entire circle that we divide by two to get the radius R.
So we calculate Pi.R^2
We do the same with the inside circle.
Then,we do the same with "the hole" in the middle.
We add the surface of the outside circle plus the surface of the " hole" that we substract from the surface of the inside circle.We obtain ananswer A.
Then we do a cross product.x being the unknown, we have
120=>surface big circle
X=> A
So x=120*A/ surface

Kim said...

The video is approximately 32 min 44 seconds long.

I took measurements as follows:
entire disk: 4cm radius
non-writeable center: 1.5cm
written portion: 0.5cm

So the area of the entire disk (including non-writeable center) is pi*(4^2) = 16pi cm^2

The area of the non-writeable center of the disk is:
pi*(1.5^2)cm = 2.25pi cm^2

-> the writeable area is 13.75pi cm^2

The written area has a width of .5cm around the edge. To get its area we subtract the area not written on from the total area
= pi * (4^2) - pi * (4-.5)^2 = 3.75 pi cm^2

So the percentage of the writeable area which has been written on is 3.75/13.75 = 27.2727%

27.2727% of the total 120 minutes = 32.7274min, or 32 minutes and 44 seconds.

Jerome said...

I don't know if I'm allowed to participate: if I am, here is my answer. If not then chuck this.

First of all, my friend has to be recording in SP mode, otherwise I have no idea what the actual time is.

Second, all I need is a rough estimate of how much has been recorded.

Third the amount recorded (in GB) is related to the area of the media recorded on (an assumption I'm not entirely sure is true). If this assumption is not true, and it depends on the groove length and space between the groves, then much more complicated methods have to be used that involve a computer and/or some sort of calculus that can figure out the total length of the tracks.

I'm assuming that a rough area estimate will do.

The fourth assumption is that I am in the United States on an American subway system.

The fourth assumption is that I have a dollar bill in my pocket, which I would not have if I was in Canada where I live, nor would our $5 do.

I need a ruler of some sort. I presume there are no kids on the subway who can supply me with one, nor is there one in my pocket. That's what the dollar bill is for. Written on every dollar is “The United States of America”, which is 28 letters in length.

I measured the radius of the DVD which is 18 letters.
The radius of the centre of the DVD to the end of the recording is L letters.
The inner radius of the DVD before the recording material is reached is 6 letters (by my measurement).

Recorded area = pi*L^2 - pi*6^2

Total recorded area = pi*18^2 – pi*6^2

Recorded area / total possible recording area = (L^2 – 36) / (18^2 – 36)

Time = Total area * 120 / total possible recorded area = 0.6 * (L^2 – 36)

The answer with all unit cancellations really is in minutes.

If I cannot use my dollar bill, I really am stuck. If I have to use a Canadian 5 (which is were I live) I am also really stuck.

PS: my birthday is actually quite close. Dec. 23 to be exact. What a yukky time to have a birthday.

Kim said...

Just realized I reversed the written and unwritten pieces in my answer -- so it's 100%-my prior answer of 27.27% = 72.72%, which makes the running time 120*72.72% or 87.27 minutes =

1 hour, 27 minutes and 16 seconds

Ilya said...

Assuming the density of data is equal in all areas of the disk, we need to calculate the written area based on the total. I will visually guess that the radius of the outside border of the written area is about 7/8th of the entire writable area's radius, while the inside border of the written area is 3/8th. The total writable area would then be Pi*(R^2-(R*3/8)^2) = Pi*R^2*55/64. The blank ring's area is Pi*(R^2-(R*7/8)^2) = Pi*R^2*15/64. So the blank area is 15/55 = 3/11 of the whole, while the written area is 8/11. Since we are guessing roughly, 120 is almost 11*11, so 8/11 of that is about 88 minutes so might as well round this to 90. P.S. It's not very intuitive that such a "narrow" strip of unwritten space represents 25% of the total, but areas of rings are funny that way :-).

TracyZ said...

For the purposes of this puzzle, 120 minutes can be contained the area of the DVD.

The area of a circle is pi*r*r (r=radius)

W: the width of the unburned portion of the DVD
with a back of envelope estimate:
the radius of the DVD from the blank center is 8W
the blank center of the DVD has a radius of 3W
the burnable portion of the DVD has a width of 5W

120 minutes = area of the DVD =
area of the circle with radius of 8W
minus the blank center of the DVD

120 minutes = ((8W*8W)*pi - ((3W*3W)*pi
120 minutes = (64W*W - 9W*W)*pi
120 minutes = 55W*W* pi
120/(55*pi) = W*W

area of unburned part of DVD
= ((8W*8W)*pi - ((7W*7W)*pi
= (64W*W)*pi - (49W*W)*pi
= 15W*W*pi

burned area of the DVD
= full area of DVD - unburned part of DVD
= 55W*W*pi - 15W*W*pi
= 40W*W*pi (then substituting for W*W from above)
= 40*(120minutes/(55*pi))*pi
= 8*120 minutes/11
= approx. 87 minutes

If Jerome's commute home (the portion when he doesn't have to drive) is 87 minutes or longer, he has enough time to watch the full movie on the DVD before reaching his house. If his commute home (non-driving portion) is shorter than 87 minutes (and I hope for his sake that it is!), then he does not have enough time to watch the full movie before getting home.

TracyZ said...

My earlier answer assumes that being the brilliant puzzle master he is, that Jerome is able to come up with his estimate of how long the movie would take to watch or at least if he does or does not have enough time to watch it, so that it doesn't reduce any potential watching time.

Maria said...

You all are brilliant and all correctly computed the length of the movie as 120 mins multiplied by the ratio of the areas of written band to the total writable band. Big fat puzzle point for anne-marie, Kim, Jerome, Ilya, TracyZ,

And happy upcoming (real) birthday to Jerome!

Tom said...

I think maybe it is not area that matters. I'm not positive but I think the rpm of the disc when playing is a constant, and if I'm right, even though the amount of data will vary a bit "from lap to lap," it is the number of laps that matters, just like a vinyl lp. If I'm right, one just measures the usable diameter and the "spent" diameter, although I am also not sure which is which. Circumference will be linear, not an issue.

And quite possibly I'm wrong, and of course a week late. There is a tremendous amount of redundant data and meta-data on a CD or DVD, which makes tracking and synchronization much better. Unlike a vinyl lp, there is a huge amount of information there beyond the movie itself. More complicated than it appears.

Anonymous said...

Its spiral

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