Max Angle v Max Rate.

[COWPAT]

Could one of you aerodynamically inclined chappies please explain to me (aerodynamically) the difference between max angle and max rate. Or maybe give me the title of some suitable reading material.

Thanks

[Matthewjharvey]

Cowpat,

Tricky without diagram but try this:

If theta is angle of climb, resolving forces parallel to direction of flight in climb :

(T-D)/W = sin(theta)

Hence best angle of climb obtained where there is maximum thrust available over that required to overcome drag. It is also best where weight is least.

Similarly, rate of climb = Vsin(theta) where V is forward speed.

From above :

Vsin(theta) = ((T-D)/W) * V

And force * velocity = power, hence

ROC = (Power available - Power Required)/W

Hence best ROC where greatest available power over power required is obtained and weight is least.

Hope this helps a little.

[COWPAT]

Thanks Greaser, now I,m really confused.

according to my O Level trigonometry. resolving forces relative to direction of flight in climb would give
sin theta = w/(T-D)

which cant be right as that would mean the heavier we are and the less thrust we have, the faster we climb.

HELP!

[sanjosebaz]

A diagram is most definitely the best way to visualise this, so you are best looking into one of many books on the subject. However, I will make a small effort to describe the rate of climb curve!... Try to draw this:

First the axes: X axis is airspeed, Y is rate of climb.
For a typical light aircraft, the curve is a kind of misshapen parabola (it's peak is shifted left of centre). Now for some numbers - draw the curve as a shifted parabola and you will get a shape near enough to be able to work with. (The real shape depends on the plane's drag characteristics and available thrust.) Make the curve fit these three points (for Cessna Skyhawk):
Y=0, X=45
Y=0, X=140
Y=700, X=80

OK, now you should have an ROC (rate of climb) curve.

Vy (best rate) is simply the speed where the curve peaks (80 knots, 700 fpm)

for Vx (best angle), draw a tangent line from X=Y=0 to touch the "parabola", then read off X and Y (actually, X is the value you are really looking for - it's the speed you are aiming for in order to clear an obstacle). On the curve I have on my lap, Vx is 69 knots (650 fpm).

Seeing the curve makes the whole thing very easy! You can clearly see that the tangent is the steepest angle you could draw on the graph and still be within the envelope.

[ 24 November 2001: Message edited by: sanjosebaz ]

[mustafagander]

From a users perspective, best RATE is min TIME to a given altitude while best ANGLE is min DISTANCE to a given altitude. :)

[Matthewjharvey]

Cowpat : to try and clarify -

Resolve in direction of flight you have Thrust acting forward, drag backward and Weight also acts against thrust with a component of Wsin(theta).

Hence by equilibrium in a steady climb the sum os all forces acting forward = sum of all forces acting rearward, thus:

T = D + Wsin(theta)

Rearranging gives

(T-D)/W = sin(theta)

Hope this clarifies a little.

Greaser

[John Farley]

COWPAT

As has been said, without diagrams it is difficult to make a convincing case.

However, if you are happy with assertions words are fine!

For maximum angle of climb, an aircraft should be flown at the speed which gives the maximum difference between thrust and drag. For piston aircraft, where thrust is reducing as speed is increased, the steepest climb will usually result from holding the unstuck speed or very close to it. For a jet aircraft, since thrust does not vary hugely with speed, the best speed is the minimum drag speed for the type. Leaving out slender deltas, min drag speeds for conventional aspect ratio aircraft could be expected to be around 200kts

On the other hand, the highest rate of climb happens when the excess power is at a maximum. For a typical piston engine aircraft this might be about 180 kt but for a jet it would be more like 400 kt.

Regards

[COWPAT]

Thanks everone for the illuminating replies, and for just plain taking the time to help me out.

All is now clear, I think.

My grateful thanks.

[THAI TUN]

In a light aeroplane would the best angleof climb be achieved with flap 0, or with a small amount of flap and a suitable small reduction in speed. Or is it type dependant?

How about a turboprop airliner, and a jet airliner?

[John Farley]

THAI TUN

Good question! I don’t really know. But admit to gut feel that a very slight improvement in angle would result if you configured and flew the aircraft as for a short field takeoff. You could always try to measure the difference! I expect Genghis can comment from experience.

As to the turbine cases again I'm not sure, but gut feel is similar. Flaps down would reduce min drag speed a tad, although total drag would be higher. However lift would be well up and again if you want to get over a (nearby) obstacle there is nothing like a lot of lift and a reduced forward speed!

Over to the experts.......

[rolling circle]

Best angle of climb, propeller or turbojet, will always be obtained in the cleanest configuration possible. Any selection of flap will result in an increase in drag. As drag increases, power required increases and the maximum rate of climb (governed by maximum excess power available) decreases. Since maximum angle of climb is a function of rate of climb/speed, if rate of climb decreases then angle of climb must also decrease.

Try drawing a graph of rate of climb against speed, in the simplest case of a propeller-driven aircraft, this will result in an igloo-shaped curve. Best angle of climb is obtained where a line from the origin is tangential to the curve. If maximum rate of climb decreases, the highest point of the curve is lower and the line from the origin is tangential at a lower rate of climb (and higher speed). The same is true for a turbojet although the curve will be much flatter.

Note that the distinction is between propeller and turbojet, not between piston and turbine.

Assuming a steady climb, lift is irrelevant since, by definition, all aereodynamic forces will remain in balance. Climb performance is all to do with power.

[Zeke]

Best angle of climb, propeller or turbojet, will always be obtained in the cleanest configuration possible. Any selection of flap will result in an increase in drag. As drag increases, power required increases and the maximum rate of climb (governed by maximum excess power available) decreases.

Incorrect, but true in the vast majority of cases, it is type dependant.

Best angle of climb occurs when the component of the thrust vector along the flight path is at a maximum. The thrust vector varies from type to type, it is not always parallel to the longitudinal and/or lateral axis (eg B747 engine thrust vector are neither parallel to the longitudinal or lateral axis), but most people simplify things and draw it parallel to the longitudinal axis.

Propeller thrust is greatest at zero airspeed and reduced with forward velocity.

The thrust is required to overcome the component of the weight along the flight path, and the drag. The other component of the weight is equal and opposite to lift, as you know T>D, and W>L for a STEADY climb.

The climb angle with flap after takeoff is type dependant. Most aircraft takeoff with flap, the aircraft gets airborne at a slower speed, but a reduced angle of climb, without flap most aircraft would take up more runway, takeoff at a higher speed, but with a better rate of climb. Looking at the net flight path after takeoff between the flap and non flap takeoff the flap takeoff in the majority of cases give a better net gradient to clear an initial obstacle.

With jets its a matter of how little fuel one uses and how much self loading or other freight is on the aircraft, sometimes a larger flap angle is used with a higher thrust setting to get a better angle, with the aircraft quickly being reconfigured and accelerated once airborne. On longer runways lower flap settings and lower thrust settings are used to take advantage of the takeoff distance available (TODA) and accelerate stop distance available (ASDA).

Best angle of climb is a performance climb and is normally only used for takeoff, with the desired aircraft configuration combination to achieve this listed in the takeoff performance charts.

The rate of climb is dependant on two factors, first the flight path, second the speed, it is the best vertical speed. It occurs when the aircraft is configured to climb at an attitude where the aircraft is configured for such that is is using the maximum excess thrust horse power. It all changes with weights, heights, configurations etc. The larger the aircraft the more detail goes into speeds, power, etc to achieve this.

The power produced by a propeller (Thrust Horsepower THP ) is equal to the engine brake HP (BHP) by the propeller efficiency.

When the aircraft is doing the most work in the vertical direction, work being W=(F * D)/T = F* (D/T) = THP*V (as V = D/T) it is flying at the attitude/airspeed combination for maximum excess power and best rate of climb.

Max angle is normally done at max thrust, best rate is normally done at max continuous power.

Z

[Fred Elliot]

Now see here, rolling circle, that answer of yours complete bilge, I say bilge.

You have confused angle with rate; exactly the point of the question. Actually, it's a very simple answer:

1. Best angle obstacle clearance) is always achieved at max power and disregards efficiency, ie drag. It may take you longer to gain a certain height (ie poor rate) but you will do it in the minimum ground distance. Since the only concern is the 'shape' of the climb, this may be grossly inefficient but probably better that hitting that water tower!

2. Best rate of climb ,ie how quickly the height is gained regardless of the distance covered, is invariably achieved at the min. drag speed or best lift/drag ratio. For a given type, this may be in a different configuration from another.

Simple, innit?

Now get back to Freshco's. ;)

[2daddies]

Simplest way to think of it:

Angle is always a lower airspeed than Rate. Therefore, less forward distance travelled and longer time to altitude but, oddly enough, greater angle.

Rate is always a higher airspeed than Angle. Therefore, more forward distance travelled and shorter time to altitude = faster rate.

Ultimately rate is a tradeoff between going fast and climbing steeply.

[rolling circle]

Zeke - Incorrect, but true in the vast majority of cases An interesting non-sequiter!

In all, however, a well reasoned argument but hardly pertinent since, as I read it, the question concerns Vx and Vy and has nothing at all to do with scheduled performance.

Fred Elliot - A thoroughly ill reasoned argument, particularly since, having rubbished my response, you proceed, in essence, to repeat it. Could it be that you are the one who is confused?

Perhaps you could drag yourself away from the all-absorbing 'Big Mac and Fries' and indicate exactly which of the following statements is "bilge, I say bilge"

1. Maximum rate of climb will be achieved at that speed where there is the greatest excess of power available over that required for straight and level flight.

2. Best angle of climb is achieved at that speed where there is the best ratio of rate of climb to speed (i.e. maximum UP for minimum ALONG).

3. On a curve of rate of climb against speed, the best ratio of rate of climb against speed is obtained where a line from the origin is tangential to the curve.

Perhaps Cowpat needs to lay down some ground rules here. After all, it is his question. Should we be talking about Vx and Vy or should we be talking about scheduled performance and Nett Take-off Flight Path? As you can see, the two are entirely different.

[COWPAT]

Oh dear, I seem to have opened a real can of worms here. All I wanted to know was the aerodynamic difference between max rate and max climb and, once I had sorted out my pythagoros, sines and cosines properly, I believe the greaser has answered it perfectly for me.

To clarify the issue. I know what max angle and max rate look like, I just wondered how they are achieved aerodynamically.

Thanks very much for the interest though everybody.