lithium2177

Hi everyone!
I live in Mexico City that is about 1600 meters high, and I fly 600 meters above it, so that would be about 7200 ft.
I've noticed that obviously because here the air is less dense than at sea level, the planes need more power. But does anyone can tell how much they are affected??
And how I can know how much more power I need here? for example I would need an engine with .10 more displacement here than at sea level? or in HP or something like that?

Thanks in advance,
Eduardo

cappio777

My Feedback: (2)

Hola Eduardo, we had a similar problem when I moved to Colorado. My regular 40 size airplanes were underpowered and I noticed that most of the guys in the club went up to the next even size engine. If your model is very light and you have marginal flight performance with the current engine first try by increasing the pitch of the propeller. In my case since my airplane was heavily loaded I replace the .46 with a .61 and it flew exactly as it did at sea level conditions. I would recommend the following engine changes:

.10 ---> .20
.20 ---> .40
.40 ---> .60

A little equation that you could use to figure out good approximation for the engine is:

Hp,s = Hp,a((rho, a)/(rho, s))^1/2

where:
Hp,s = horsepower at sea level
Hp,a = horsepower at altitude
rho,s = standard air density at sea level
rho,a = standard air density at that altitude

espero esto te ayude un poco

lithium2177

Thanks for the advice, and for the formula.
I ll try to use it.

Muchas gracias, si me va a ayudar bastante!.

Eudardo

Jimmbbo

FWIW,

A rule of thumb is that a normally aspirated full scale engine develops about 75% of its sea level power at full throttle at 7500 feet in the standard atmoshpere (7500 ft density altitude)... Would expect model engines to do the same...?? [:-]

Best,

Jim

Jack Hyde

My Feedback: (1)

Why is there a sqrt in the density ratio? The volume of air taken into the combustion chamber is independent of the density. The mass is directly proportional to density. I thought power would decrease directly with atmospheric density.

cappio777

My Feedback: (2)

Ooops, its called a typo, just reverse the a and the s on the HP.
Thank you for the correction.


Hp,a = Hp,s((rho, a)/(rho, s))^1/2

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