Cold Weather More HP?

[Topless]

Originally Posted by blue2000s

Topless, it's true that a turbine engine, such as a jet on an airplane can also make more power under cooler air conditions, but power isn't the reason for the longer runway needed for take-off. That's due to the differences in ar density providing differences in lift. Hotter air, being less dense, provides less lift so the plane needs to be going faster before the lift force is sufficient to raise the plane.

Yeah I know... which is exactly what I said.

[rick3000]

I never actually thought about it, but now that you mention it. I have been feeling more power since it has gotten colder. Thanks for pointing that out.

It would be interesting to see a dyno test at 70° and at 35° to see if there is a significant power difference.

[blue2000s]

Originally Posted by Topless

Yeah I know... which is exactly what I said.

The way I read it, it sounded like you were referring to variations in engine power having something to do with the length needed for take off.

[ichiro75]

Here's the verdict. 1% increase in Horsepower every 10 degree colder. :dance: You do the math. It makes quite a difference. Comparing 90-100 degree summer vs. the 10-20 degree winter.

http://blogs.sportcompactcarweb.com/1004173/editorials/cold-weather-horsepower/index.html

http://image.sportcompactcarweb.com/f/editorials/cold-weather-horsepower/1032071+w700+cr1+re0+ar1/dyno-pull.jpg

[blue2000s]

Originally Posted by ichiro75

Here's the verdict. 1% increase in Horsepower every 10 degree colder. :dance:

It's not quite that much, but it's a decent rule of thumb.

[efahl]

Assuming you are talking about Fahrenheit degrees, and only density effects, it's actually bigger than 1%/10F. The math is quite trivial, just turn your temperatures into absolute (Rankine scale) and divide them (the ideal gas law, more specifically Guy-Lussac's law, only have linear terms making things easy).

E.g.,

t0 = 70F = 529R
t1 = 60F = 519R
density change = t0/t1 = 529/519 = 1.019

In other words, 1.9% increase in density.

(Hey, admins, fix the [ code ] block in your php...)

[blue2000s]

Originally Posted by efahl

Assuming you are talking about Fahrenheit degrees, and only density effects, it's actually bigger than 1%/10F. The math is quite trivial, just turn your temperatures into absolute (Rankine scale) and divide them (the ideal gas law, more specifically Guy-Lussac's law, only have linear terms making things easy).

E.g.,

t0 = 70F = 529R
t1 = 60F = 519R
density change = t0/t1 = 529/519 = 1.019

In other words, 1.9% increase in density.

(Hey, admins, fix the [ code ] block in your php...)

The % will be different depending on the actual temperature (40 vs 50 vs 60 ect) but that's just density, not power.