Cold Weather More HP?

[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.