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v0.31 Talk:Temperature
Conversion corrections
The conversions are faulty, if you should give many temperatures to konvert to, you should see to that the conversions are the same at more than one point...
--Zorbeltuss 10:55, 25 June 2010 (UTC)
Or rather the same at any point concerning the Kelvin and Rankine scales...
--Zorbeltuss 10:59, 25 June 2010 (UTC)
- The Kelvin and Rankine scales are not directly equivalent. Kelvin scale has the same degree interval as Celsius, while Rankine scale has the same degree interval as Fahrenheit. Nevertheless, the Rankine conversion equation is wrong, so I'll fix it. --DeMatt 11:28, 25 June 2010 (UTC)
- Ah, I put it badly, as I didn't know the exact numbers to be put I couldn't decide if the number to add from Rankine or Kelvin were right, so I couldn't fix it my self, the multiplier was correct though... --Zorbeltuss 16:03, 8 July 2010 (UTC)
- the numbers seem to be correct now though--Roderik 15:03, 11 August 2010 (UTC)
- The conversions are still faulty, the conversion between Celsius and Kelvin is (C + 273 = K), so if the Celsius temperature is [DF scale] = [CELSIUS] * 9/5 + 10000 than Kelvin can't be [DF scale] = [KELVIN] * 9/5 + 9508.33. It should be [DF scale] = [KELVIN] * 9/5 + 9727 --Hypoaktivna ovca 15:45, 27 October 2010 (UTC)
- The original was correct: C * 9/5 + 10000 = (K - 273.15) * 9/5 + 10000 = K * 9/5 - (273.15 * 9/5) + 10000 = K*9/5 - 491.67 + 10000 = K * 9/5 + 9508.33. One way to verify this is that zero Kelvin should give the same result as zero Rankine, since they are the same temperature; likewise 273.15 Kelvin and zero Celsius. --Neil 13:39, 29 October 2010 (UTC)
- The conversions are still faulty, the conversion between Celsius and Kelvin is (C + 273 = K), so if the Celsius temperature is [DF scale] = [CELSIUS] * 9/5 + 10000 than Kelvin can't be [DF scale] = [KELVIN] * 9/5 + 9508.33. It should be [DF scale] = [KELVIN] * 9/5 + 9727 --Hypoaktivna ovca 15:45, 27 October 2010 (UTC)
- the numbers seem to be correct now though--Roderik 15:03, 11 August 2010 (UTC)
- Ah, I put it badly, as I didn't know the exact numbers to be put I couldn't decide if the number to add from Rankine or Kelvin were right, so I couldn't fix it my self, the multiplier was correct though... --Zorbeltuss 16:03, 8 July 2010 (UTC)
May I point you to another error? It is just a small one, and has nothing to do with conversion from one scale to another. The absolute zero (zero Kelvin) is the temperature that causes the molecules of an ideal gas to stop completely. Therefore, it is only the absolute zero point for an ideal gas. So, in theory, it is actually possible to reach temperatures below 0 Kelvin. -- Amras 14:59, 23 January 2011 (UTC)