Your pet peeve, technical or otherwise.

[Kim Christensen]

Did you read that sentance before you quoted it or was it just a handy one to paste in?

Getting a little hot under the collar? Looks like you are up about 50%...   

[Analog Kid]

Because 0V = 0V.  0A=0A, 0Hz = 0Hz

0*C does not equal "no heat".

What is 20% hotter than 0*C?

What is 20% more current than 0A?

Surprised nobody bit this:
It's zero, of course.

[Analog Kid]

If there are no objections here, I think we can safely say:

QED: Percentages are not valid when used with [non-absolute] temperature values.

[paulca]

But, can you say it is 20% hotter without mentioning units?

</troll>

On the topic of thermometers and extreme environments.  Physics does provide a temperature for the vacuum.  Of course it's all QED and in K.

But it's an interesting puzzle to try and wrap your mind around.  If you tried to measure the temperature in space with various devices what kind of results do you think you would get?  You are free to fix the shading vs. sunlit or explore it too.

[TimFox]

But, can you say it is 20% hotter without mentioning units?

</troll>

On the topic of thermometers and extreme environments.  Physics does provide a temperature for the vacuum.  Of course it's all QED and in K.

But it's an interesting puzzle to try and wrap your mind around.  If you tried to measure the temperature in space with various devices what kind of results do you think you would get?  You are free to fix the shading vs. sunlit or explore it too.

Only if you are using absolute temperature:  20% hotter works for both Kelvin and Rankine.

[themadhippy]

so wots the difference between 9 below zero and  3 degrees

[TimFox]

I think we all can agree that both the general population and physicists understand the concept of "degrees" in temperature, although some may need to remember the formulas for conversions between scales.

[CatalinaWOW]

I think any mention of temperature without units is prone to misunderstanding.  I was reminded of this as I was reading a non-fiction book by UK author (he is ambiguous over whether he is British or Irish) born sometime around 1950.  The book is published in 2014 and he regularly describes the temperature as a few degrees below zero and as below freezing.  What units is he using.  Based on his imprinting from his birth year it might be Fahrenheit, but publication date might imply Centigrade.  His emphasis on how cold it was could be because he is using Fahrenheit or it could be because he was under dressed and has poor cold tolerance.  The location he is describing easily supports either hypothesis.

Even non-technical communication depends on clarity.

[Circlotron]

When watching Facebook videos full screen on my iPad using Brave browser. When the start button is pushed it mutes the audio, and when the audio is subsequently unmuted it pauses the video… Endless loop. 😡

[paulca]

When watching Facebook videos full screen on my iPad using Brave browser. When the start button is pushed it mutes the audio, and when the audio is subsequently unmuted it pauses the video… Endless loop. 😡

I don't  use facebook anymore, but I still do use Messenger.

Facebook however, since I am using Brave browser which keeps recycling my browser UID is freaked out because it can't track me.  So every single time I log in to messenger it throws a security alert email and asks me for an email code etc.  Then redirects me to facebook to try and refresh the tracking cookies.

I love that it's torturing it.  I love how obviously desparate it is for those juicy UUID cookies and... no.. it can't have them.

[calzap]

Incorrect use of “percent increase”.   Should be later value minus earlier value divided by earlier value times 100, i.e. change related to starting value.   So, if value was 100 and increased to 200, percent increase is (200-100)/100*100 = 100.  But can find it as (200-100)/200*100 = 50 percent increase.  Or as 200/100*100 = 200 percent increase (this one is commonly used for investment sales).  Or even (200-100)/[(100+200)/2]*100 = 67 percent increase.

Mike

So, what is your take on marketing claiming a % increase/decrease in temperature?

Gamer's Nexus went absolutely nuts at some manufacturers for saying things like a case had 15% better cooling and 5% lower temperatures.

He was like, YOU CANNOT APPLY PERCENTAGE TO A DEGREE!!!! and basically detonated.

Oh, you can correctly calculate a % increase or decrease in temperature.  Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.   That 50C object is 122F.  A 50% decrease from 122F is 61F.  But 25C is 77F, not 61F!  A percentage increase or decrease in an intensive variable with an arbitrary zero isn’t all that quantitatively useful by itself.   There can be directional and ordinal comparisons.   A decrease is colder; increase is warmer.  Comparing two identical objects that are cooling from the same starting temperature, the one with the biggest % decrease in temperature got colder.

Calcs get a little messy if a % change takes the temperature through zero.  If an object is 16C and has a 50% decrease in temperature, that’s 16 x 0.5 = 8.  What about a 150% decrease?  100% decrease would take it to zero.  A further 50% decrease takes it to 16 x -0.5 = -8.

However, % decrease/increase in temperature can be used in calculations with other quantities to obtain useful results.   If the mass of an object, it’s specific heat, it’s starting temperature and % decrease/increase in temperature are known, then a potentially useful % change in its thermal energy content can be calculated.

Percentage changes in cooling can have useful meaning depending on how it’s calculated.  If 100 J per second was being removed from a source and now it’s 150 J per second, that’s a 50% increase in cooling, i.e. 50% better cooling.

Mike

[Analog Kid]

Oh, you can correctly calculate a % increase or decrease in temperature. Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.

Well, of course that's been shown (by @TimFox) to be wrong.
Yes, you can decrease the number representing the temp by 50%, but that doesn't bring the actual temperature down by that amount. The only way that works is if you use an absolute temp scale (like Kelvin).

Or is that what you're trying to get across here?

I'm convinced by TimFox's arguments.

[helius]

Many (even most?) formulas involving temperature do not refer to absolute temperature, but relative temperature difference (∆T). With temperature differences, K and °C are the same unit.

[calzap]

Oh, you can correctly calculate a % increase or decrease in temperature. Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.

Well, of course that's been shown (by @TimFox) to be wrong.
Yes, you can decrease the number representing the temp by 50%, but that doesn't bring the actual temperature down by that amount. The only way that works is if you use an absolute temp scale (like Kelvin).

Or is that what you're trying to get across here?

I'm convinced by TimFox's arguments.

Yes, that is what I was showing too.  If you start at equivalent temperatures in the C and F scales, and decrease the number by 50%, the two results are not equivalent.  But if used correctly, the percentage decrease or increase can lead to useful results regardless of scale.  And no, it doesn't have to be an absolute scale.

Mike

[Analog Kid]

I don't see how they can be useful if they're meaningless. The only accurate thing you can say about two temperatures on a non-absolute scale is "X is (colder/hotter) than Y".

[TimFox]

Oh, you can correctly calculate a % increase or decrease in temperature. Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.

Well, of course that's been shown (by @TimFox) to be wrong.
Yes, you can decrease the number representing the temp by 50%, but that doesn't bring the actual temperature down by that amount. The only way that works is if you use an absolute temp scale (like Kelvin).

Or is that what you're trying to get across here?

I'm convinced by TimFox's arguments.

Yes, that is what I was showing too.  If you start at equivalent temperatures in the C and F scales, and decrease the number by 50%, the two results are not equivalent.  But if used correctly, the percentage decrease or increase can lead to useful results regardless of scale.  And no, it doesn't have to be an absolute scale.

Mike

I submit that percentage changes in non-absolute temperature are never useful.
However, you can compare temperature differences.
If percentages were useful in temperature readings, why are they never used in practice by careful writers?

[calzap]

Oh, you can correctly calculate a % increase or decrease in temperature. Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.

Well, of course that's been shown (by @TimFox) to be wrong.
Yes, you can decrease the number representing the temp by 50%, but that doesn't bring the actual temperature down by that amount. The only way that works is if you use an absolute temp scale (like Kelvin).

Or is that what you're trying to get across here?

I'm convinced by TimFox's arguments.

Yes, that is what I was showing too.  If you start at equivalent temperatures in the C and F scales, and decrease the number by 50%, the two results are not equivalent.  But if used correctly, the percentage decrease or increase can lead to useful results regardless of scale.  And no, it doesn't have to be an absolute scale.

Mike

I submit that percentage changes in non-absolute temperature are never useful.
However, you can compare temperature differences.
If percentages were useful in temperature readings, why are they never used in practice by careful writers?

Specific heat of iron is 0.45 J/g/deg C.   A kg piece of iron was 200 C, and its temperature dropped 50%.  How much thermal energy did it lose?
Answer:  1000 * 0.45 * 200 * (1 – 50/100) = 45,000 J

[TimFox]

That’s a correct calculation for a drop of 100 C^o.
What would the calculation give for a “50%” drop from 392^o F or 473 K?
What is the physical difference between 200^o C and the other two temperatures?

[Ice-Tea]

No, honey, you didn't just 'literally' die.

[CatalinaWOW]

Oh, you can correctly calculate a % increase or decrease in temperature. Question is whether it has any meaning.  Suppose an object is 50C.  A 50% decrease in temperature brings it to 25C.

Well, of course that's been shown (by @TimFox) to be wrong.
Yes, you can decrease the number representing the temp by 50%, but that doesn't bring the actual temperature down by that amount. The only way that works is if you use an absolute temp scale (like Kelvin).

Or is that what you're trying to get across here?

I'm convinced by TimFox's arguments.

Yes, that is what I was showing too.  If you start at equivalent temperatures in the C and F scales, and decrease the number by 50%, the two results are not equivalent.  But if used correctly, the percentage decrease or increase can lead to useful results regardless of scale.  And no, it doesn't have to be an absolute scale.

Mike

I submit that percentage changes in non-absolute temperature are never useful.
However, you can compare temperature differences.
If percentages were useful in temperature readings, why are they never used in practice by careful writers?

Specific heat of iron is 0.45 J/g/deg C.   A kg piece of iron was 200 C, and its temperature dropped 50%.  How much thermal energy did it lose?
Answer:  1000 * 0.45 * 200 * (1 – 50/100) = 45,000 J

If you know what you are doing, percentage change in temperature can be used to get a correct answer.  But if you don't fully understand what is going on it is easy to completely screw things up.  And even if you do get it right without understanding the limitations, it doesn't provide the basis for solving more general problems.

This reminds me of a test back in university days, when I didn't understand Smith charts (still don't grok them).  The question involved matching a couple of stubs in a waveguide system.  I plotted a set of points and gave an answer which proved to be correct.  Prof asked me to explain why the answer was correct, I couldn't.  I was just desperately trying to come up with something.  He didn't understand why it had worked either.  So the success of the calculation was of no use to any other problem.

[TimFox]

We technical people often dislike “bean counters”, but they do know percentages.
For a loan or bond yield, a change from 3% to 4% is + 1 “percentage point”.
Stating that as +33% would be confusing.
For a stock market index, an increase from 10,000 to 10,500 is “+ 5%”.

[calzap]

That’s a correct calculation for a drop of 100 C^o.
What would the calculation give for a “50%” drop from 392^o F or 473 K?
What is the physical difference between 200^o C and the other two temperatures?

1000*0.45*473*(1-50/100) = 106,425 J

Different than a 50% drop from 200 C, of course.  Point is that if you are given a starting or ending temperature and % change, a useful calculation can be done.

Mike

[TimFox]

Not a useful calculation, just a confusing one.
A different answer in energy for apparently the same change.

[Analog Kid]

Well, it would work if you converted the temps to be compared to an absolute scale and then did the comparison.

Otherwise no.

[rsjsouza]

IMO it is not foreign the idea of saying that "today the temperature dropped to half of what it was yesterday" - i.e., 50% less.
Sure, in a typical conversation no one would come up with additional fraction numbers such as 25 or 33%, but that does not invalidate the use of percentages for temperatures.