how do these effect flight?
#1
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From: Sarasota,
FL
my friend and i got into a discusion, were trying to figure out how these characterictics affect flight and lift and speed on our model aircraft
low altitude compared to high altitude (we fly at 8000' msl, so the air would be thinner correct? - but do we need to go faster to lift off?)
hot days vs cold days
low humidity vs high humidity
and how does barametric pressure relate to humidity? or does it all?
thanks
mainly im interested in what these things do to speed and lift
thanks
low altitude compared to high altitude (we fly at 8000' msl, so the air would be thinner correct? - but do we need to go faster to lift off?)
hot days vs cold days
low humidity vs high humidity
and how does barametric pressure relate to humidity? or does it all?
thanks
mainly im interested in what these things do to speed and lift
thanks
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From: Los Angeles,
CA
Anything that decreases the density of the air (like high altitude, low barometric pressure and higher temperature) will give worse lift and less power from the engine. However, it will also reduce drag (a bit!).
The effects of high humidity are much more debatable.
Being at 8000 feet will make a very noticable difference as the pressure is much less and will have the above effects.
-David C.
The effects of high humidity are much more debatable.
Being at 8000 feet will make a very noticable difference as the pressure is much less and will have the above effects.
-David C.
#4
there is lots of solid data on this - some is quite weighty.
Here is a short version
the lower the effective density - the more lift is needed.
The easy proof is in using an electric motor powered model
At sea level- you can set a base line for watts required to fly in a fixed manner.
Now get in a balloon and launch at various higher altitudes.
Read how many watts are used to do the same tests as you go higher .
The big difference will be the changes needed in the prop used (increased pitch and diameter ) to draw the same watts.
You will need to go faster to get same lift and if you can "hover" you will see the prop requirements change as the air gets thinner ( more pitch and diameter to hold the same watt loading.
The plane can go faster the higher you get as there is less density to swim thru -. On an electric, the motor (power source) is constant at any altitude - so the result will be easy to see.
you can go faster on the same power .
On a gas powered model - the power decreases faster than the reduction in drag so you can't get a valid comparison.
Here is a short version
the lower the effective density - the more lift is needed.
The easy proof is in using an electric motor powered model
At sea level- you can set a base line for watts required to fly in a fixed manner.
Now get in a balloon and launch at various higher altitudes.
Read how many watts are used to do the same tests as you go higher .
The big difference will be the changes needed in the prop used (increased pitch and diameter ) to draw the same watts.
You will need to go faster to get same lift and if you can "hover" you will see the prop requirements change as the air gets thinner ( more pitch and diameter to hold the same watt loading.
The plane can go faster the higher you get as there is less density to swim thru -. On an electric, the motor (power source) is constant at any altitude - so the result will be easy to see.
you can go faster on the same power .
On a gas powered model - the power decreases faster than the reduction in drag so you can't get a valid comparison.
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From: Los Angeles,
CA
ORIGINAL: briansommers
how does increasing the temperature make the air less dense??
are there less molucules in hot air then cold? and if so why and how?
how does increasing the temperature make the air less dense??
are there less molucules in hot air then cold? and if so why and how?
This is the principal behind an intercooler in an engine. If the air going towards the inlet is hot but is passed through a heat exchanger to cool it down before actually going into the cylinder, more air gets into the cylinder (it's denser) and more power is produced.
-David C.
#6
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Water vapor is less dense that air so higher humidity makes air less dense, and also decreases relative oxygen content. It is actual humidity, not relative humidity, which is important.
Jim
Jim
#7
Brian,
Designers often use a "Standard Atmosphere" to account for the effects of altitude. This works pretty well when two things happen: 1) the temperature at sea level is about 59 degrees Fahrenheit, and 2) the temperature decreases by about 3 1/2 degrees F for every 1,000 ft increase in altitude (standard lapse rate). On a "standard day", the density at 8,000 ft MSL is only about 79% of the density at seal level. From this you can estimate the airspeed needed for takeoff at 8,000 feet to be about 1.13 times the speed needed for takeoff at sea level (assuming you use the same rotation technique).
To account for hot days, there is also a standard "hot day" atmosphere (where the temperature at sea level is 103 degrees F). On a "hot day" at 8,000 ft, the density is only 74% of what it would be on a standard day at sea level (the temperature at 8,000 ft would be 73 degrees F). This would make the takeoff speed 1.16 times the standard day sea level takeoff airspeed.
On a standard day with 100% relative humidity at 8,000 ft, the density is 99.7% what it would be with 0% relative humidity at 8,000 ft, so you can see that humidity doesn't change things very much at all (Jim is absolutely right, it is the actual humidity that is important, but the equation I'm using converts to actual humidity, and I think it's easier to think in terms of relative humidity).
I'm not too current with these calculations, so someone else out there might want to check my work. I hope this helps.
Designers often use a "Standard Atmosphere" to account for the effects of altitude. This works pretty well when two things happen: 1) the temperature at sea level is about 59 degrees Fahrenheit, and 2) the temperature decreases by about 3 1/2 degrees F for every 1,000 ft increase in altitude (standard lapse rate). On a "standard day", the density at 8,000 ft MSL is only about 79% of the density at seal level. From this you can estimate the airspeed needed for takeoff at 8,000 feet to be about 1.13 times the speed needed for takeoff at sea level (assuming you use the same rotation technique).
To account for hot days, there is also a standard "hot day" atmosphere (where the temperature at sea level is 103 degrees F). On a "hot day" at 8,000 ft, the density is only 74% of what it would be on a standard day at sea level (the temperature at 8,000 ft would be 73 degrees F). This would make the takeoff speed 1.16 times the standard day sea level takeoff airspeed.
On a standard day with 100% relative humidity at 8,000 ft, the density is 99.7% what it would be with 0% relative humidity at 8,000 ft, so you can see that humidity doesn't change things very much at all (Jim is absolutely right, it is the actual humidity that is important, but the equation I'm using converts to actual humidity, and I think it's easier to think in terms of relative humidity).
I'm not too current with these calculations, so someone else out there might want to check my work. I hope this helps.
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From: Holts Summit, MO
Brain,
Here is a link you might want to look at:
[link]http://wahiduddin.net/calc/density_altitude.htm[/link]
[link]http://wahiduddin.net/calc/cf.htm[/link]
Steve
Here is a link you might want to look at:
[link]http://wahiduddin.net/calc/density_altitude.htm[/link]
[link]http://wahiduddin.net/calc/cf.htm[/link]
Steve




