Climb Performance and Weight

[uwochris]

Hey guys,

I have been ratting my brain out trying to figure out a proper way to explain the relationship between weight and climb performance, in terms of nose attitude.

Question: for any given IAS, how does weight affect the nose attitude to maintain in the climb?

My argument: if weight increases, the amount of excess power available will be less. This implies the climb angle and rate will be less steep. If the same nose attitude was maintained (i.e. the same attitude compared to an exactly similar a/c at half the weight), the AOA would increase due to the fact that the flight path has become less steep (i.e. if the same attitude is maintained but the flight path is made less steep, the AOA must increase). Because the AOA increases, the IAS will decrease due to the increase in induced drag. To return to the original IAS, the nose must be lowered, which implies the climb angle will be further reduced. In the end, the heavier a/c will have a less steep climb angle, a lower rate of climb, and a lower nose attitude compared to another a/c of lesser weight.

Is this correct?

I also tried to explain it this way: if weight increases, that means that to achieve a balanced climb, the amount of lift will have to increase. The pilot can increase lift by changing the AOA or by changing the IAS. This meanst that for any IAS in a climb, the AOA must be higher. The AOA will adjust as the flight path changes and becomes less steep. To maintain the previous IAS, the nose must be lowered further to compensate for the increased induced drag. The heavier a/c flying at the exact same IAS will have a lower nose attitude, but a higher AOA.

This seems like such a simple concept, but I just don't know how to go about explaining the effects of weight on the nose attitude, for any IAS.

Thanks in advance for any responses,

Chris.

 

[roundout]

My argument: if weight increases, the amount of excess power available will be less.

Is this correct?

Chris.

nope. weight doesn't affect the bhp which the engine can put out.

 

[sstearns2]

OK, I'll give this a try.



Rate of climb (ROC) is purely true airspeed (TAS) and flight path angle (FPA), meaning the angle between the flight path of the airplane and the horizon. FPA is the angle you see on your attitude indicator minus your angle of attack.



ROC = Sin (FPA) * TAS * (6000/60)

The 6000/60 changes knots to ft/min.


So if you in a 172 at 75 Knots TAS and a flight path angle of 5 degrees above the horizon you get a rate of climb is 654 ft/min, independent of the weight of the airplane or if it even has an engine.



The question then becomes can the airplane maintain 75 knots at a flight path angle of 5 degrees. This is where weight and excess horsepower come into play.



I could go on with the physics and trig if you'd like, but for most people I'd probably just use a car going up a hill analogy.



Scott

 

[midlifeflyer]

I assume you're not looking for a physics lesson. I think you are essentially correct.

if weight increases, the amount of excess power available will be less. After all you're using the same horsepower to move a lighter object.

And that's reflected in the airspeeds that will produce your best rate of climb. The "book rate" for Vy, for example, is based on max gross. Like most other aircraft speeds, Vy decreases with a decrease in weight and increase with an increase in weight. And an increase in airspeed is usually reflected in a lowering of the nose.

 

[SDCFI]

How bout trying it this way? The higher weight req's a higher AOA for a given airspeed to create the same amt of lift. The higher AOA results in higher induced drag. Thrust is the opposite of drag, so the increased drag results in lower excess power, therefore a lower climb rate.

 

[DGdaPilot]

How bout trying it this way? The higher weight req's a higher AOA for a given airspeed to create the same amt of lift. The higher AOA results in higher induced drag. Thrust is the opposite of drag, so the increased drag results in lower excess power, therefore a lower climb rate.

Excess power or excess thrust? Doesn't excess power relate to horsepower? Maybe I'm mistaken.

 

[JetBlast2000]

Clarification

For lift and weight to be in equilibrium in order to maintain any desired attitude of flight, more lift must be produced to balance the heavy weight. To achieve this, the airplane must be flown at an increased angle of attack. As a result, the wing will stall sooner (i.e. at a higher airspeed) when the airplane is fully loaded than when it is light. Stalling speed in turns (that is, at increased load factors) will also be higher. In fact, everything connected with lift will be affected. Take-off runs will be longer, angle of climb and rate of climb will be reduced and, because of the increased drag generated by the higher angle of attack, fuel consumption will be higher than normal for any given airspeed. Severe g-forces are more likely to cause stress to the airframe supporting a heavy payload.

When we talk about Thrust we are talking about a Force generated by the powerplant which imparts a change in velocity of mass. F= M A. Thrust and excess thrust are functions of airspeed. You have the most excess thrust at Vx. This gives us a climb over a unit of distance. The total drag curve = thrust required. Power is a unit of work done over a specific time. The most excess power occurs at Vy. Where we have the greatest climb per unit of time.
As has been stated, weight does not affect what the engine can produce.

 

[sstearns2]

I've been meaning to get back to this.....

OK, I'll give this a try.

Rate of climb (ROC) is purely true airspeed (TAS) and flight path angle (FPA), meaning the angle between the flight path of the airplane and the horizon. FPA is the angle you see on your attitude indicator minus your angle of attack.





ROC = Sin (FPA) * TAS * (6000/60)

The 6000/60 changes knots to ft/min.



So if you in a 172 at 75 Knots TAS and a flight path angle of 5 degrees above the horizon you get a rate of climb is 654 ft/min, independent of the weight of the airplane or if it even has an engine.





The question then becomes can the airplane maintain 75 knots at a flight path angle of 5 degrees. This is where weight and excess horsepower come into play.

I think a little digression into how max rate of climb is measured when testing an airplane would help. It might seem like you would just read the rate of climb off the VSI at various altitudes and airspeeds, but there is a much faster and more elegant way of doing it. Let's say we're going to measure max rate of climb at 3000 feet. You level off at 3000 feet and slow the airplane to just above stall. Now you go to full power, maintain 3000 feet and start a timer. As the airplane accelerates you record the airspeed every few seconds (obviously a data acquisition system helps a lot). You end up data from which you can figure out how fast the airplane is accelerating at any airspeed at 3000 feet. Hold that thought.



Gravity is an acceleration, 32.2 feet per second, per second within a couple hundered miles of the surface of the earth. If you drop a feather and a bowling ball next to each other in a perfect vacuum they will fall right next to each other and they'd be traveling at 32.2 feet per second after one second, 64.4 feet per second after two seconds and so on.



Let's go back to our 5 degree Flight path angle example above. Gravity is trying to decelerate the airplane back down that 5 degree slope, but the engine has enough excess power to match it and you end up with a constant airspeed climb.



Gravitational deceleration = 32.2 * Sin (FPA)



= 2.8 feet/sec^2 or 1.7 Knots/sec at a FPA of 5 degrees **



So, the airplane must be able to accelerate in level flight at a minimum of 1.7 Knots/sec at 75 knots and 3000 feet in order to be able to climb at a FPA of 5 degrees nose up at 3000 feet at a constant 75 knots.



The numbers I’ve used roughly equal Cessna 172 performance. So, the next time your in a 172 level off and slow to 65 knots. Go to full power and maintain altitude. Start a timer at 70 knots and stop it at 80 knots. If it takes about 6 seconds then the airplane is accelerating at average 1.7 knots per second. Therefore, it will be able to maintain a constant airspeed climb at 75 knots at a FPA of 5 degrees nose up with full throttle, which we figured above to be about 650 feet/minute.



Weight, or more correctly mass, comes into play because a heavier 172, or more massive one, will not accelerate as well as a light one, so your ability to oppose gravitational decceleration in a climb is reduced.

** At a FPA of 90 degrees or straight up you need to oppose the full brunt of gravity (The Sin of 90 degrees is 1). So you need an airplane that will accelerate at 32.2 ft/sec^2 or 19 knots/second in level flight in order to climb straight up at a constant airspeed. So, next time your in your F-15......

Hope this helped,

Scott

 

[sstearns2]

The nice thing about looking at it in the above way is that is really doesn't matter if you are in a C-172 or an F-15 or even a Honda Accord for that matter.


Scott