XKCD: metric conversions

[iago]

http://xkcd.com/526/

I declare that guy an idiot (or somebody who lives somewhere warm).

Right now where I live, it's "Spit goes 'clink' cold", and that's a slightly cold winter day (and my spit doesn't go 'clink', I've tried). 

-5 = cold day in Moscow? Ha! If they're anything like here, -30 is a standard winter day. :)

</rant>

[Newby]

I thought it was funny. I guess if you don't use the metric system it's funny. :P

[iago]

It IS funny, but their temperatures sound like they were written by somebody from CA or FL, they needed a Canadian to go over them and tell them what "Cold" means. :)

[Towelie]

So, to a person from CA (aka me), it makes perfect sense. And I also just spent over an hour on xkcd lol.

[iago]

So, to a person from CA (aka me), it makes perfect sense.

But it's wrong!

-5 isn't a cold day is Moscow, and -10 isn't a cold day in Boston (or maybe it is). And -20 isn't FUCKFUCKFUCKCOLD. They need to bump everything about -15 - -20 lower. :)

[Hitmen]

But it's wrong!

-5 isn't a cold day is Moscow, and -10 isn't a cold day in Boston (or maybe it is). And -20 isn't FUCKFUCKFUCKCOLD. They need to bump everything about -15 - -20 lower. :)

I think the guy lives in Boston, but -10 is rather cold for here. We don't drop below 0 very often, our coldest days usually hover around it. Unless you are talking about celsius in which case I don't know what -10 is, but considering 0 = 32 it is probably still above 0F so not too terribly cold. :)

[Sidoh]

OMFG AN ARTICLE OF HUMOR IS INACCURATE!

[iago]

I think the guy lives in Boston, but -10 is rather cold for here. We don't drop below 0 very often, our coldest days usually hover around it. Unless you are talking about celsius in which case I don't know what -10 is, but considering 0 = 32 it is probably still above 0F so not too terribly cold. :)

It IS Celcius, that's the whole point of the comic. :)

A "cold day in boston" is 0F, according to you, which is ~-20C, thus proving my point. :D

OMFG AN ARTICLE OF HUMOR IS INACCURATE!

The whole point of xkcd's humour is that it's generally accurate/factual. :P

I can suspend disbelief for the sake of humour, but not random inaccuracies! :P

[Sidoh]

Wwaaaaaaaaaaaaahhhhhhhh

[zorm]

Right now where I live, it's "Spit goes 'clink' cold", and that's a slightly cold winter day (and my spit doesn't go 'clink', I've tried). 

Clearly the problem here is that you aren't tall enough to provide the water sufficient time to freeze.

Lets say you spit 1 gram of water and its at the same temperature as your body (~37C)

So dQ = M*Cp*dT + M*L = 1/1000 * 4.18 * 37 + 1/1000 * 334000 = 334.15 J

So if you spit from a height of 2m you have 0.64 seconds before it hits the ground. Thus you need to remove ~523.03 J s^-1 to freeze the water before it hits the ground.

After this you can go ahead and compute how much energy the wind is removing from the air per second and such.

[iago]

So, one of you math guys needs to calculate how long it takes a sphere with roughly the density of water to freeze!

I seem to recall a formula like pv/nrt for this kind of thing. Ring any bells? :)

Also, what if I spit at a 45 degree angle upwards, and manage to get it, say, 10 feet? That means it's spending significantly more time in the air.

[Towelie]

So, one of you math guys needs to calculate how long it takes a sphere with roughly the density of water to freeze!

I seem to recall a formula like pv/nrt for this kind of thing. Ring any bells? :)

Also, what if I spit at a 45 degree angle upwards, and manage to get it, say, 10 feet? That means it's spending significantly more time in the air.

Do you know what would be faster than using equations? :P

going outside and trying it ;)

[deadly7]

Right now where I live, it's "Spit goes 'clink' cold", and that's a slightly cold winter day (and my spit doesn't go 'clink', I've tried). 

Clearly the problem here is that you aren't tall enough to provide the water sufficient time to freeze.

Lets say you spit 1 gram of water and its at the same temperature as your body (~37C)

So dQ = M*Cp*dT + M*L = 1/1000 * 4.18 * 37 + 1/1000 * 334000 = 334.15 J

So if you spit from a height of 2m you have 0.64 seconds before it hits the ground. Thus you need to remove ~523.03 J s^-1 to freeze the water before it hits the ground.

After this you can go ahead and compute how much energy the wind is removing from the air per second and such.

IN THIS HOUSE, WE OBEY THE LAWS OF THERMODYNAMICS.

[Blaze]

Add a "LISA, " and you're all set!

[iago]

So, one of you math guys needs to calculate how long it takes a sphere with roughly the density of water to freeze!

I seem to recall a formula like pv/nrt for this kind of thing. Ring any bells? :)

Also, what if I spit at a 45 degree angle upwards, and manage to get it, say, 10 feet? That means it's spending significantly more time in the air.

Do you know what would be faster than using equations? :P

going outside and trying it ;)

I already said, it doesn't work! I've tested in at least -40 or so, although we may dip below that tonight (I've heard rumours we might be hitting -50 (including windchill, of course)).