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Sep 6th, 2007, 10:41 PM
#1
Thread Starter
Lively Member
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Sep 7th, 2007, 12:58 AM
#2
Re: Temperature question (help!)
For (a) I can help,
It's not exactly described in the same method, but it works just as good.
Lets look at the difference between 3500 and 2000.
It's 1500.
1500 is 75% of 2000.
Now the decline in temperature for 2000 was 10.
You take 75% from 2000 and also 75% from 10; resulting to 1500 and 7.5.
You subtract 7.5 from 3, and there you have the -4.5 which is the answer.
Delete it. They just clutter threads anyway.
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Sep 7th, 2007, 01:15 AM
#3
Re: Temperature question (help!)
 Originally Posted by Yunie
T
Working solution:
(a) Temperature difference = 3 - 13
= -10 That should be clear!
Now you know that for a climb of 2000m the temperature decreases be 10°
That would make a decrease of (10/2000)° per m of climb.
If you want to climb to 3500m just multiply that by 3500!
3500*(-10/2000)° = (3500/200)*(-10)
For a height of 3500m, temp difference = (3500/2000) * (-10)
= -17.5
Now you know that for a climb of 36500m the temperature would decrease by 17.5°, since the actual temperature at sealevel is 13°, just subtract 17.5° from that one!
Temp is 13 - 17.5 = -4.5 degree celsius.
(b)
The actual Sealevel temperature is 13°, so the temperature difference down to the dangerous temp is -17° -13°, which is -30°
Temp. difference = -17 - 13
= - 30
Now you need to calculate the difference in altitude you would need top get a decrease of 30°. Knowing that for each meter of climb the temperature decreases by (10/2000)° , an the inverse would say that (2000/10) m of climb are needed to get a decrease of 1°.
You need 30° decrease, so it is 30*(200/10) = (30/10)*2000
maximum height = (30/10) * 2000
= 6000m
BTW: This "dangerous temperature" is nonsense! Severe icing can only happen close to the freezing point (0°),since in temperatures below that, there would be no humidity in the air. It's f****** cold, but not dangerous!
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Wait, I'm too old to hurry!
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Sep 7th, 2007, 05:07 AM
#4
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