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Climbing at -6 degrees vs 0 degrees?

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WmInce   
11 minutes ago, Ed Cesnalis said:

:) They [airliners] are accepting more drag for a shorter take off roll, it doesn't produce a steeper angle once they are climbing nor does it produce extra stall protection it simply lowers stall speed.

Lift is a constant. They cannot accept more drag without using more power to equal the increased drag amount. With more power used to equal more drag less power is left to seek best climb angle, angle is reduced as a result of the take off flaps.  

You can't budget more power for the dirty configuration and somehow have more power left for a steeper climb instead you have less power left and a more limited climb angle.

I think you are confusing or mixing climb angle with climb rate. 

During climbout, at Vx, a lower climb rate is sacrificed for a better climb angle. IOW, it is steeper angle, which covers less distance over the ground, for the amount of altitude gained. Whereas, Vy goes for best rate (altitude gained for a given time period).

Vx is a speed which results in an altitude gain in a minimum distance covered.

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ctfarmer   

There is nothing clean about a high angle of attack. That is pure and simple, drag inducing. At -6, your angle of attack will be higher to achieve the same rate of climb at the same speed compared to 0.

I say, you might be surprised at the outcome.

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Just now, WmInce said:

I think you are confusing or mixing climb angle with climb rate. 

During climbout, at Vx, a lower climb rate is sacrificed for a better climb angle. IOW, it is steeper angle, which covers less distance over the ground, for the amount of altitude gained. Whereas, Vy goes for best rate (altitude gained for a given time period).

Vx is a speed which results in an altitude gain in a minimum distance covered.

No I understand the difference between Vx and Vy.

Flaps produce drag that results in a reduction in available thrust used to obtain Vy or Vx or even best speed. Therefore best rate, best angle and best speed are all reduced with more flaps.

I was confused for decades on this too and the source of confusion comes from using take of flaps to clear obstacles. Its needed, it works, it looks like the heavy's but its because of the shorter take of roll. The take off flaps produce a short take off roll that produces the best close in obstacle clearance, we know that and we assume its steeper but its less steep.

599c748442fce_15vszero.jpg.c0f2739ff39f8a7c32a49e01e6e329d4.jpg

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Imagine a helicopter taking off with forward speed of 5kt, and a climb of 300fpm.  Now imagine a CTSW with 900fpm climbing at 78kt climbing out beside it.  At the end of the 2000ft runway, which will be at the higher altitude?  The helicopter will, because even though it's climb is much less, it has far more TIME to climb because of the slower speed before reaching the trees.

Now back to the CT, for which I just did this math:

At -6°, a CT climbs at 78kt (89.7mph) at 900fpm (15fps).  On a 2000ft runway, it takes him 15.2 seconds to reach the trees at the end, at which time he's at 228 feet.  

Another CT climbs at 15° at 60kt (69mph) at 700fpm (11.66fps).  At the end of the same 2000ft runway, he's at 231ft, 3ft higher.  Not a big deal, unless the trees are 230ft tall.  ;)

Because the height is greater at the same distance, and the geometry is a right triangle, the angle of climb in the 15° setting MUST be larger.  The -6° CT gets 15.2 seconds of climb at 900fpm to clear the trees.  The 15° CT gets 19.8 seconds of climb at 700fpm to clear them.  The extra 4.6 seconds of climb, even at a lower rate, lead to a higher total climb angle.  

I'm with Bill and Tom here.  I think Ed is discounting/disregarding time of climb before the obstacle in all this.      

 

 

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3 minutes ago, ctfarmer said:

There is nothing clean about a high angle of attack. That is pure and simple, drag inducing. At -6, your angle of attack will be higher to achieve the same rate of climb at the same speed compared to 0.

I say, you might be surprised at the outcome.

Your right that a high angle is drag inducing. 

Your wrong on the rest.  Don't confuse deck angle with angle of attack.

 

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1 hour ago, Ed Cesnalis said:

Yes your probably right that vertical speed with -6 can be greater than zero and still be shallower but that's not the point.  The point is that the cleaner the configuration, like -6 over zero or zero o over 15 or -6 over 15, the steeper Vx will be.  

Vx at -6 is steeper than Vx at zero or Vx  at 15 because in both cases less drag results in more available thrust to use seeking best angle.

Your first two sentences contradict each other. Your last sentence is wrong because the excess power doesn't necessarily have a direct relation to angle of climb. Excess power does have a direct effect on rate of climb. With 15° flaps you do have more drag and less available power for climb, there is no disputing that. The percentage of reduction in climb rate due to the lower excess power is offset by an even greater reduction in percentage of forward speed. That is how you have a steeper angle.

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9 minutes ago, FlyingMonkey said:

I just did this math:

At -6°, a CT climbs at 78kt (89.7mph) at 900fpm (15fps).  On a 2000ft runway, it takes him 15.2 seconds to reach the trees at the end, at which time he's at 228 feet.  

Another CT climbs at 15° at 60kt (69mph) at 700fpm (11.66fps).  At the end of the same 2000ft runway, he's at 231ft, 3ft higher.  Not a big deal, unless the trees are 230ft tall. 

This is not valid.  You are using made up numbers and leaving out the ground roll component.  At my field you cannot get airborne at -6 in 2,000' but it takes 400' at 15 degrees.

The short takeoff roll is what makes the 15 degree departure better at clearing obstacles close in and you don't include take off rolls in your math.

 

9 minutes ago, FlyingMonkey said:

Because the height is greater at the same distance, and the geometry is a right triangle, the angle of climb in the 15° setting MUST be larger.  The -6° CT gets 15.2 seconds of climb at 900fpm to clear the trees.  The 15° CT gets 19.8 seconds of climb at 700fpm to clear them.  The extra 4.6 seconds of climb, even at a lower rate, lead to a higher total climb angle.  

What is a total climb angle?  The -6° CT was in the air a much shorter distance and had to climb much steeper to arrive at the same point within 2'.

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JLang   
36 minutes ago, Ed Cesnalis said:

No I understand the difference between Vx and Vy.

 

I agree with WmInce, I think you are confusing Vy and Vx.  Your sketch even mixes them up, since you show Vy when discussing clearing an obstacle.

Also, lift is not constant between flap configurations.  More flaps = more lift.  This is why liftoff is sooner with flaps.  Of course, drag is also increased, but the point that I think you are missing is that for best angle, at Vx, this is beneficial.

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27 minutes ago, Ed Cesnalis said:

This is not valid.  You are using made up numbers and leaving out the ground roll component.  At my field you cannot get airborne at -6 in 2,000' but it takes 400' at 15 degrees.

The short takeoff roll is what makes the 15 degree departure better at clearing obstacles close in and you don't include take off rolls in your math.

 

What is a total climb angle?  The -6° CT was in the air a much shorter distance and had to climb much steeper to arrive at the same point within 2'.

Disagree.  These numbers are assuming that both airplanes begin their climb from zero at the same point on the runway, 2000' from the obstacle.  If anything, it is skewed toward the performance at -6° because it does not account for time to accelerate to the higher climb speed at -6°.  

These are not "made up numbers" they are about what I see from my airplane at the two flap settings.  

By "total climb angle" I simply meant the true angle of climb when you take all the factors (including time to climb) into account.  How do you reconcile that a higher flap setting might have a slower climb rate but more time to make the climb before reaching the obstacle? (Time x climb rate = height at obstacle), but you only seem to want to talk about climb rate, and ignore the time component.

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2 minutes ago, Tom Baker said:

Your first two sentences contradict each other. Your last sentence is wrong because the excess power doesn't necessarily have a direct relation to angle of climb. Excess power does have a direct effect on rate of climb. With 15° flaps you do have more drag and less available power for climb, there is no disputing that. 

My first 2 sentences are in agreement they point out that just because the -6 can be less steep, which is true it doesn't prove any point because it isn't less steep when angle of attack is set to best rate of climb.

Excess power doesn't necessarily have a direct relation to the angle of climb your right.  We assume when we compare best rate at the different flap settings that the pilot then sets the angle of attack to achieve the best rate of climb.  If the pilot does nothing then you guys could be right but inelegant argument requires the pilot to use one of 3 pitch attitudes to compare performance.  The pilot has to pitch to Vx to compare best angle or he has to pitch to Vy to compare best rate or he has to pitch to level flight to compare best speed.  In any of the 3 comparisons -6 will win as long as you don't include ground roll.

-6 wins in all 3 cases for the same reason, less drag requires less thrust and leaves more available thrust to seek angle or  rate or speed. This is only true if the pilot sets the angle of attack to best rate, or best angle, or best speed.  If the pilot uses the same stick position in each comparison instead of optimizing the angle of attack then all bets are off.

26 minutes ago, Tom Baker said:

The percentage of reduction in climb rate due to the lower excess power is offset by an even greater reduction in percentage of forward speed. That is how you have a steeper angle.

Really?

Again when seeking to clear the obstacle using 15 the pilot should pitch for best rate of climb.  The pilot adjusts how much of the loss of excess power comes from rate and how much comes from forward speed.  The pilot can adjust his angle but his best angle will be less than the best angle at -6, he won't be able to achieve that due to having less available thrust.

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11 minutes ago, WmInce said:

Ed,

Are you assuming Vx is a set value for all flap settings?

 

Good question Bill.  I am assuming that Vx is the best angle at the current configuration.  I can't rely on my POH it only has one number and doesn't say for which setting.

 

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WmInce   
29 minutes ago, WmInce said:

Ed,

Are you assuming Vx is a set value for all flap settings?

 

16 minutes ago, Ed Cesnalis said:

I am assuming that Vx is the best angle at the current configuration.  I can't rely on my POH it only has one number and doesn't say for which setting.

The reason why I ask that . . . is because Vx actually varies with each respective flap or airfoil configuration change. On more sophisticated aircraft, that value is tagged and will slide up and down the airspeed tape, as the airfoil changes shape.

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7 minutes ago, FlyingMonkey said:

Disagree.  These numbers are assuming that both airplanes begin their climb from zero at the same point on the runway, 2000' from the obstacle.  If anything, it is skewed toward the performance at -6° because it does not account for time to accelerate to the higher climb speed at -6°.  

These are not "made up numbers" they are about what I see from my airplane at the two flap settings.  

Andy, I know you are a smart guy but your observations are 2' different and that isn't even observable.  Plus the rotation points or climb initiation points are hugely different between 15 and -6.  For me 15 means 400' and -6 means thousands of feet. I have trouble seeing how you can even observe such a comparison in your own flying.

 

7 minutes ago, FlyingMonkey said:

By "total climb angle" I simply meant the true angle of climb when you take all the factors (including time to climb) into account.  

Okay, angle is angle, 2 components distance and time.  Total means true when factoring in time?  Lets let this one go if its not important.

18 minutes ago, FlyingMonkey said:

How do you reconcile that a higher flap setting might have a slower climb rate but more time to make the climb before reaching the obstacle? (Time x climb rate = height at obstacle), but you only seem to want to talk about climb rate, and ignore the time component.

Is this a trick question?  A slower rate but more time is reconciled by lower forward speed / steeper angle?

I'll be glad to talk about the time component but I'm not sure what you want me to address.

 

 

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7 minutes ago, WmInce said:

 

The reason why I ask that . . . is because Vx actually varies with each respective flap or airfoil configuration change. On more sophisticated aircraft, that value is tagged and will slide up and down the airspeed tape, as the airfoil changes shape.

yup

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8 minutes ago, Ed Cesnalis said:

Andy, I know you are a smart guy but your observations are 2' different and that isn't even observable.  Plus the rotation points or climb initiation points are hugely different between 15 and -6.  For me 15 means 400' and -6 means thousands of feet. I have trouble seeing how you can even observe such a comparison in your own flying.

 

Okay, angle is angle, 2 components distance and time.  Total means true when factoring in time?  Lets let this one go if its not important.

Is this a trick question?  A slower rate but more time is reconciled by lower forward speed / steeper angle?

I'll be glad to talk about the time component but I'm not sure what you want me to address.

 

 

My example was just that, an example.  The goal was to show that climb rate can be lower and still lead to a higher altitude at the obstacle.  Higher altitude at obstacle means the climb angle is greater.  Since Vx is best angle of climb, in the example Vx is greater at 15° than at -6°.   It's just geometry, I'm not trying to trick you!  :)

 

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Energy Budget

L = power needed to offset weight

AP = Available power

T = min thrust needed at -6,  x = additional thrust needed for zero,  y = addtional thrust needed for 15 degrees, z = additional thrust needed for 30 degrees.

  Lift thrust to equal drag available for best angle climb
30 degrees 1,300lb T+x+y+z Available Power - L - (T+x+y+z)
15 degrees 1,300lb T+x+y Available Power - L - (T+x+y)
0 degrees 1,300lb T+x Available Power - L - (T+x)
-6 degrees 1,300lb T Available Power - L - T

More flaps means more drag and less power available to devote to achieving the best angle of climb.

Works the same way for best rate or best speed.

 

 

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Just now, FlyingMonkey said:

My example was just that, an example.  The goal was to show that climb rate can be lower and still lead to a higher altitude at the obstacle.  Higher altitude at obstacle means the climb angle is greater.  Since Vx is best angle of climb, in the example Vx is greater at 15° than at -6°.   It's just geometry, I'm not trying to trick you!  :)

 

The problem is your example shows the result you want but not why.  

Vx is not greater at 15° than at -6° because of the huge drag difference Vx is far less at 15° than at -6°.

A lot of power is required to offset the additional drag at 15 and not much poop is left compared to -6 to devote to the best climb angle.

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7 minutes ago, Ed Cesnalis said:

The problem is your example shows the result you want but not why.  

Vx is not greater at 15° than at -6° because of the huge drag difference Vx is far less at 15° than at -6°.

A lot of power is required to offset the additional drag at 15 and not much poop is left compared to -6 to devote to the best climb angle.

It shows *exactly* why:  Because the airplane at 15° is climbing slower, which gives more time in the climb before reaching the critical point (at the obstacle).  The fact that the -6° airplane in the example is climbing 200fpm faster is not enough to offset the additional 4.6 seconds of climb the 15° airplane gets.

990fpm = 15fps

700fpm = 11.66fps

 

to travel 2000ft takes 15.2sec at 78kt for the -6° airplane

to travel 2000ft takes 19.8sec at 60kt for the 15° airplane

 

15fps x 15.2sec = 228ft

11.66 x 19.8 = 230.86ft

 

The time component overwhelms the climb rate component, though only slightly in this example.  But once you concede this is possible, it's apparent that the faster climbing airplane does not always have the advantage in Vx.

 

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1 hour ago, Ed Cesnalis said:

My first 2 sentences are in agreement they point out that just because the -6 can be less steep, which is true it doesn't prove any point because it isn't less steep when angle of attack is set to best rate of climb.

Really?                                                                                                                                                                                                                                                      ANGLE                                                                                                                                     Again when seeking to clear the obstacle using 15 the pilot should pitch for best rate of climb.  The pilot adjusts how much of the loss of excess power comes from rate and how much comes from forward speed.  The pilot can adjust his angle but his best angle will be less than the best angle at -6, he won't be able to achieve that due to having less available thrust.

Excess power only effects rate of climb. Angle of limb is determined by rate over distance. A lower rate over an even lower distance provides a steeper angle. The pilot adjust the speed to give the best climb rate balanced against a lower forward speed to clear the obstacle. The speed he uses to clear the obstacle will not produce the best overall rate of climb.

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WmInce   
23 minutes ago, Ed Cesnalis said:

. . . Vx is not greater at 15° than at -6° because of the huge drag difference Vx is far less at 15° than at -6°.

A lot of power is required to offset the additional drag at 15 and not much poop is left compared to -6 to devote to the best climb angle.

In the first line above, it seems your are treating Vx as if it is "an angle." It is not. It is a speed to achieve best angle of climb (max altitude gained for a given distance). And that value changes as the properties of the airfoil changes.

With Vx, climb rate is irrelevent.

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27 minutes ago, FlyingMonkey said:

The time component overwhelms the climb rate component, though only slightly in this example.  But once you concede this is possible, it's apparent that the faster climbing airplane does not always have the advantage in Vx.

I never said that the faster climbing airplane always has the advantage in Vx but I am saying that your airplane has its fastest Vx at -6.

I think what we observe on our departures is the result rotation speed and proximity to obstacles far more than actually getting a feel for 15 Vx vs -6 Vx.  Who does -6 departures anyway?  If they are not common knowing the relative angle at all isn't likely.

I have done 4 neg6 departures this week all at gross weight. 2 from Mammoth and 2 from sea level but all 4 with 7,000'+ runways, no obstacles and downhill and I will admit that I didn't have the knowledge about the benefits of -6 departures from 15 degree starts until I have done a number of them.  Kind of like flying in box canyon's, I'm doing that over 100 times a year now while I am working so I have a feel that I never had before.

In closing I enjoy your posts but think your method in this case doesn't produce usable results for comparing Vx rates.  As pilots we learn to compare the result of ground roll, acceleration and Vx rates and their combined impacts on our departures which led to us all thinking, me included that climb was steeper with flaps when it isn't.

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42 minutes ago, Tom Baker said:

Excess power only effects rate of climb.

Sorry Tom but that's just silly.  Excess power / Available Thrust is there and used at the pilots discretion.

I can and do put my nose on the horizon and my throttle to the forward stop and in doing so I am applying all Excess power to obtain best speed.

I can and do put my nose at a higher pitch attitude where my rate of climb is maximized and in doing so I am applying all Excess power to obtain best rate of climb.

I can and do put my nose at a higher pitch attitude where my angle of climb is maximized and in doing so I am applying all Excess power to obtain best angle of climb.

I'm free to use settings any where between Vso and Vne.

Excess power effects what I tell it to by where I set the angle of attack.

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1 minute ago, Doug G. said:

Pages 5-8 through 5-13 in the CTLS POH give roll and takeoff distances for flap settings and weight.

Can you paste them?  do they seam correct?

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