3 ms·

But not really. 100°C times two is not 200°C. 100°C times two is 473°C

ethbr0 3y ago
Howso? If the conversion to Kelvin is additive (as opposed to something more complicated), and multiplying in Kelvin is fine, then how is multiplying effectively-unsigned in Celsius not fine?
- sgjohnson 3y ago
  Ok, another example. Is 10°C 10 times as hot as 1°C? No. Multiplying kelvin is fine because kelvin is an absolute scale.
  - _a_a_a_ 3y ago
    > No. For a given baseline, perhaps yes?. Kelvin has its baseline inherent, for C you can pick what you like if you're clear about it. Perhaps. (Edit: and using only positive or negative temps as pointed out above)
    - ethbr0 3y ago
      Yes. Parent's spaghetti examples weren't the most illustrative. What I believe they're griping about is that "10°C does not have 10x as much heat as 1°C." In actuality, "10°C has 10x the difference in heat that 1°C has relative to the freezing point of water." So it's a comparison definition quirk, not a numerical one.
- yarg 3y ago
  c = k + 273 x * c = x * k + x * 273 This leaves multiplied celsius off by a delta of (x - 1) * 273.
  - ethbr0 3y ago
    That's showing multiplied Celsius in terms of Kelvin, no? 1 Celsius = 274.15 Kelvin 10 Celsius = 283.15 Kelvin Obviously 283.15 K != 10 * 274.15 K. But that's not what's being said, per my lower comment -- it's 10x the difference relative to the freezing point of water, compared to 1°C. The original upthread comments I were taking issue with were saying that multiplication has no meaning / is invalid in Celsius. It does have a meaning, and it's mathematically consistent. It may not mean what's being assumed, but nuance doesn't mandate "thou shalt never," IMHO.
    - yarg 3y ago
      When I say multiply by n, what I'm really doing is applying a delta of (n - 1) times my distance to the origin. The origin with celsius isn't at zero, so you're prevented from doing what you were pretending was multiplication (at least in the most natural way of doing it). Obviously you can negate your offset, multiply and reapply your delta. It's a bit damned awkward though.

Howso? If the conversion to Kelvin is additive (as opposed to something more complicated), and multiplying in Kelvin is fine, then how is multiplying effectively-unsigned in Celsius not fine?

sgjohnson 3y ago
Ok, another example. Is 10°C 10 times as hot as 1°C? No. Multiplying kelvin is fine because kelvin is an absolute scale.
- _a_a_a_ 3y ago
  > No. For a given baseline, perhaps yes?. Kelvin has its baseline inherent, for C you can pick what you like if you're clear about it. Perhaps. (Edit: and using only positive or negative temps as pointed out above)
  - ethbr0 3y ago
    Yes. Parent's spaghetti examples weren't the most illustrative. What I believe they're griping about is that "10°C does not have 10x as much heat as 1°C." In actuality, "10°C has 10x the difference in heat that 1°C has relative to the freezing point of water." So it's a comparison definition quirk, not a numerical one.
yarg 3y ago
c = k + 273 x * c = x * k + x * 273 This leaves multiplied celsius off by a delta of (x - 1) * 273.
- ethbr0 3y ago
  That's showing multiplied Celsius in terms of Kelvin, no? 1 Celsius = 274.15 Kelvin 10 Celsius = 283.15 Kelvin Obviously 283.15 K != 10 * 274.15 K. But that's not what's being said, per my lower comment -- it's 10x the difference relative to the freezing point of water, compared to 1°C. The original upthread comments I were taking issue with were saying that multiplication has no meaning / is invalid in Celsius. It does have a meaning, and it's mathematically consistent. It may not mean what's being assumed, but nuance doesn't mandate "thou shalt never," IMHO.
  - yarg 3y ago
    When I say multiply by n, what I'm really doing is applying a delta of (n - 1) times my distance to the origin. The origin with celsius isn't at zero, so you're prevented from doing what you were pretending was multiplication (at least in the most natural way of doing it). Obviously you can negate your offset, multiply and reapply your delta. It's a bit damned awkward though.