Filling up Winthrop Lake with discs

[grodney]

How many discs would you have to throw into the Lake at Winthrop to raise the water level 1 inch?

(I hope several of you work it out before looking at ANY of the spoilers. p.s. I didn't actually measure the volume of a disc....do it all with assumptions and known tech standards.)

Multiple choice:

Spoiler

A) 17
B) 1,700
C) 170,000
D) 17,000,000

Assumptions:

Spoiler

Disc: Flight plate, excluding rim, is 17.3cm diameter and 1.5mm thick. Rim is a 1.5cm equilateral triangle that is 19.2cm long (19.2 is average of outside diameter of 21.1cm and inside diameter of 17.3cm).
Lake: 37,905 square meters minus 1,020 square meters for the island, with vertical walls at the shore, and is entirely enclosed.

Answer:

Spoiler

D) ~17 million discs

Disclaimer:

Spoiler

It's Friday, I might have messed up the math. The 1.5mm flight-plate thickness is a guess, but hopefully gets us within a couple orders of magnitude. Corrections are welcomed, as are challenges to any of the assumptions.

 

[joshhyzer]

Someone is bored.

 

[grodney]

Someone is bored.

 

[sillybizz]

Spoiler

17 million seems low

 

[tbird888]

No assumption that all discs rest flightplate down?

 

[filobedo]

How many discs would you have to throw into the Lake at Winthrop to raise the water level 1 inch?

(I hope several of you work it out before looking at ANY of the spoilers. p.s. I didn't actually measure the volume of a disc....do it all with assumptions and known tech standards.)

Multiple choice:

Spoiler

A) 17
B) 1,700
C) 170,000
D) 17,000,000

Assumptions:

Spoiler

Disc: Flight plate, excluding rim, is 17.3cm diameter and 1.5mm thick. Rim is a 1.5cm equilateral triangle that is 19.2cm long (19.2 is average of outside diameter of 21.1cm and inside diameter of 17.3cm).
Lake: 37,905 square meters minus 1,020 square meters for the island, with vertical walls at the shore, and is entirely enclosed.

Answer:

Spoiler

D) ~17 million discs

Disclaimer:

Spoiler

It's Friday, I might have messed up the math. The 1.5mm flight-plate thickness is a guess, but hopefully gets us within a couple orders of magnitude. Corrections are welcomed, as are challenges to any of the assumptions.

Next assignment should be who can figure out how PDGA ratings are calculated.

 

[grodney]

No assumption that all discs rest flightplate down?

It doesn't matter, does it?

 

[tbird888]

If there's air trapped under the flightplate, I'd say that makes a difference. If you assume that there's zero potential for air being trapped, regardless of how the discs rest, it doesn't.

 

[grodney]

If there's air trapped under the flightplate, I'd say that makes a difference. If you assume that there's zero potential for air being trapped, regardless of how the discs rest, it doesn't.

Yeah I thought of that so I guess my assumption was no air trapped because then they would float....but you're right, a little air could be trapped, so I guess there's another assumption that no air is trapped.....and probably that the discs aren't consumed into mud at the bottom or something.....i.e. assume pure displacement of plastic=water.

 

[tbird888]

Thank you for the clarification. I wasn't talking about a full amount of air the entire height of rim or anything. That would definitely keep the disc above water.

 

[brutalbrutus]

That is a pretty substantial lake... so I would say all of them...

 

[BogeyNoMore]

Goose crap will raise the water level in that lake by an inch loooooong before plastic does.
That said, goose poop is bio degradeable, plastic ain't. :\

 

[sillybizz]

I know it's not a creek but maybe donkey_punched would throw his bag in. :|

We need to pass a law that says all chuckers must play this course. We'll get to 17 million at the end of the Summer, and about 170 million beer cans. :|

 

[BogeyNoMore]

I went at it thusly, having not looked at any of the spoilers.

Basically, you'd have to cover the bottom of the lake with a layer of discs that has a net volume that equates to displacing an inch of water.

• Estimating surface area of the lake at 448,000 sq ft , or 64,512,000 sq in, you'd need 64,512,000 cu in of discs (very crudely measured with google maps, because I can't download googele eath without permission I wasn't about to ask for).

• Given the shape of a typical disc, I'd estimate their volume to be approx 1/3rd that of a cyl.
• Volume of a cyl = π r² h
• With an avg diameter of 21.5cm, avg rim depth of 1.3cm… I estimate volume of an avg disc to be about: 9.6 cu in.
• At 9.6 cu in per disc, it would take 6,719,744 discs to total 64,512,000 cu in, thus raising the water level by an inch.


Curious to know what Grodney has disc volume at to come up with his answer.

 

[grodney]

I had 408,000 sq ft, minus 11,000 sq ft for the island. So my lake is, what, 10% smaller than yours.

For the disc, I had 54 cu cm, compared to your 157 cu cm. So your disc is 3 times the volume.

Thanks for doing that!

 

[tbird888]

Answer unnecessary

 

[BogeyNoMore]

I didn't subtract the area of the island. Thought about it, but figured as a source of error, it wasn't as great as the other estimates I'd made.

 

[markmcc]

Another way to look at disc volume is to look at the disc weight. If we accept that an "average disc" will float at around 140 grams, and sink at 150 grams (no trapped air), can we accept that a disc at 145 grams will just barely sink? If so, that disc will displace 145 grams of water, which is 145 cubic centimeters. If you don't like the 145 gram number pick your own, but I think that we're easily within 10% now.

Way easier than trying to measure the volume of a disc...

 

[DG_player]

Archimedes is rolling in his grave!

The only proper way to measure the volume of a disc is to drop it in a bucker of water.

 

[markmcc]

I agree. To get it exactly you'd have to measure the weight of water displaced and then convert it to volume.

But think about it. If the disc "just" sinks, then the equivalent specific gravity of the disc is 1.0 (or 1.00001). If the specific gravity is 1.0 then the density of the disc is the same as the density of the water. Therefore the weight of water displaced by the disc equals the weight of the disc...