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Question regarding best rate and best angle of climb.

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Question regarding best rate and best angle of climb.

Old 1st Apr 2020, 14:26
  #1 (permalink)  
Thread Starter
Join Date: Jun 1999
Location: Wherever the work is...
Posts: 123
Question regarding best rate and best angle of climb.

Hi all,

I understand that Vx results in best climb angle and that this occurs only at the speed at which there is maximum excess thrust.

Likewise, I also understand that Vy results in best climb rate and that this occurs only at the speed at which there is maximum excess power.

my question is this:
why is it that climb angle if a function of thrust, and climb rate is a function of power?
Why is it not the other way around?

I feel like I'm missing something important here, and would just like to understand why thrust relates to climb angle and not rate, and why power relates to climb rate and not angle.

cheers all for any help!
Gearupandorrf is offline  
Old 2nd Apr 2020, 08:06
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Join Date: Sep 2016
Location: USA
Posts: 575
Thrust and angle are more basic, simple relationship. Take that as the baseline. Excess thrust yields climb angle.

Now, remember that power is thrust times speed.

Then, take thrust and angle, and multiply them both times speed. Thrust becomes power (from the above law). And angle becomes rate. Not really exactly a multiplication, but... a certain steepness of angle, with a higher speed than before, gives a higher climb rate than before. Itís proportional.
Vessbot is offline  
Old 2nd Apr 2020, 09:23
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Join Date: Aug 1999
Location: England
Posts: 1,028
to explain it you need some Physics, which you may or may not have. I don’t kniw.

Go and research:
Work = force x distance
Power = force x speed
Power = energy / time
gravitational potential energy
Power required
Power available
Excess power

and you’ll see why rate of climb relates to power (specifically, excess power)

Climb angle is a bit easier:
research the definitions of the orientation of the forces on an aircraft
look to see how the forces relate to one another in a climb
you will see that the geometry of the forces is related to thrust (specifically, excess thrust)
Capt Pit Bull is offline  
Old 8th Apr 2020, 20:09
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Join Date: May 1999
Location: UK
Posts: 35
Ok, take a PA-28 and stand it on its tail pointing straight up. Put an engine and prop combination up front that can generate enough thrust (or force) to overcome the aircraft weight (plus any drag generated when it starts to move). Voila! Best angle of climb = 90 degrees and only requires that thrust exceeds weight and drag. The Americans experimented with this in the 50's with vertical take off prop aircraft (google Convair XFY).
The higher you climb, the more work is done because: work = force times distance. However, the rate of climb in terms of height gained in a certain amount of time could be pitifully slow.
To get a half decent rate of climb you will need an excess of power because power = force times distance (work done) divided by time. Or more specifically, an excess of power available compared to power required.
Does that help?
Black Jake is offline  

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