Flying Instructors & ExaminersA place for instructors to communicate with one another because some of them get a bit tired of the attitude that instructing is the lowest form of aviation, as seems to prevail on some of the other forums!

I don't know if this is the right place to put this topic but i need some explanation. I don't seem to understand if you climb at vx you will gain an altitude in the shortest distance and if you climb at vy you gain the most altitude in the least time.

If you climb at vx and you reach 2000' in, let's say, 5nm while in the mean time you reach in 5nm at vy, 1300' why in the hell are going to climb at Vy to gain altitude in the shortest time?????

I've been looking up these things but they are always refering to excess power for Vy and excess thrust for Vx. Why not taking excess power into account for Vx.(too theoretically for me)

I simply do not understand, who can explain a bit?

The simple vertical profile of a normal cross country flight (or every other one at that) is: takeoff, climb, cruise, descent and landing. So, since aviation isn't free (hope to see a day when Avgas/Jet A-1 will be free), you want to fly as many nautical miles per minute/hour you can. In order to do that, you need to have as high ground speed as you can, but if you want to save some fuel, you should fly higher - and to fly higher you need to climb.

The optimum between vertical and horizontal speed during climb is usually (C172/PA28/C152/....) Vy, Vy+5 or Vy+10, it depends on conditions... Climbing with Vy instead of Vx also gives you additional pros: engine cooling is much (MUCH) better since the forward speed is higher, margin from Vx to stall is usually only 10-15 knots, which means that if you get a 10 knot wind shear or forgot to check the ASI for few seconds, it wouldn't be nice... Also, most AFM/POH (especially Cessna's) say that climbing at Vy or higher gives you better forward visibility, which has proven useful many times in VFR where we follow "see and be seen" rule...

To extend the issue to IFR: which climb would expedite the traffic departing behind you - climb where you would be at a "safe" level (level at you must be in order that succeeding aircraft can depart) 2 NM from airport in 10 minutes or climb where you would be at the same level 15 NM from airport in 3 minutes? Do the math and it will come to you.

But still, if you need scientific explanations, there are million of books available which you can study...

Think of it in terms of "not hitting the ground!" Gradient is the means to avoid terrain (or achieves minimum SID levels etc) whereas Rate gets you to more efficient or useful levels quicker, assuming that the gradient it gives is adequate to avoid terrain. If not, adjust speed to achieve the gradient required and accept the prolonged climb period. As they say "better to be late in this life than early into the next!"

The best rate of climb speed Vy will cause one to gain the most height in a given time period.

Imagine you are taking off with a 610ft hill 1nm ahead.

If the aircraft climbs at 60Kt (Vy) and tghis gives a rate of climb = 600ft/min the aircraft will be 600ft at the hill and will impact 10 ft below the top.

The whole idea of flying at the speed for best angle is to sacrifice a small amount of climb rate in return for a (proportionally) bigger reduction in speed.

Let's say that we now fly the same example at the best angke speed of 50Kt(Vx). This gives us a reduced rate of 550ft per minute.

However, even though we are climbing at a reduced rate it now takes longer to travel the 1nm to the hill and you can simply work it out that we will now

travel 5/6nm in 1 minute (climb 550ft)

and then climb 110ft in the remaining time it takes to travel the 1/6nm left to the hill.

Therefore despite climbing at a reduced rate we are now at 660ft as we safely pass over the hill.

Think of these V speeds as tools you can use as a pilot..

Vx: Your flying along, you look forward and see a big hill up ahead...Vx is the speed in which you take to gain the most altitude in a given distance...and yes it is the steepest climb angle...but it's about obstacle clearance....you only have some much distance to get over this thing...in a small plane this speed might be 60kts...

Vy is more about the relationship between forward speed and vertical speed....meaning if your goal for example is to get up to altitude the quickest...you use Vy...now Vy on this same little plane might be 70 kts...it's a quicker speed then Vx, because speed translates into time, time can translate to FPM...and when it comes to getting up to alt...Max FPM is what we are chasing....

Another way to look at Vx vs Vy...imagine your need to get out of a canyon and there is no way you just fly straight out of it, so you do a circling climb, effectively making no forward progress to your destination......you could use Vy under these circumstances as that will give you the lest amount of time to get up and over...

Vx is a trade off of FPM and forward distance traveled, trading it for a higher climb angle vs Vy...

Hello D boy-you must be wondering why nobody is answering your questions. I have highlighted your questions to make it easier for the experts

Quote:

Hello

I don't know if this is the right place to put this topic but i need some explanation.

I can answer this one--Probably not, because as yet no one on here has found the answer on the internet. (these types of question are outside the knowledge of most instructors)

Quote:

I don't seem to understand if you climb at vx you will gain an altitude in the shortest distance and if you climb at vy you gain the most altitude in the least time.

The question is why?

Quote:

If you climb at vx and you reach 2000' in, let's say, 5nm while in the mean time you reach in 5nm at vy, 1300' why in the hell are going to climb at Vy to gain altitude in the shortest time?????

The question is in bold

Quote:

I've been looking up these things but they are always refering to excess power for Vy and excess thrust for Vx. Why not taking excess power into account for Vx.(too theoretically for me)

Quote:

I simply do not understand, who can explain a bit?

Tx

The question is in bold

So D boy, has anyone answered your question(s) yet?

The purpose of this forum is to exchange information within the flying training sphere. If all you want to do is take cheap shots at instructors while providing no value added to this topic what so ever, than I strongly suggest you go find another audience to bother.

Thrust is a force, and power is a rate at which work is done. Work is the product of force times the distance moved in the direction of the force.

Vx is the speed at which you get the best angle of climb. Climbing is a matter of overcoming the force of gravity. Therefore the more force (excess of thrust over drag) you have, the better you can do it (ingoring all other considerations). So thrust is a key issue

Vy is the speed at which you get the best rate of climb. Rate of climb is connected to rate of delivery of work, and therefore power is a key issue.

Vy is the lowest point of the drag curve. Any other speed (including Vx if not the same) will have higher drag.

Hope that helps.

Last edited by puntosaurus; 11th May 2010 at 21:40.

I'm not an FI, but by chance I was reading up on exactly this today.

Excess power matters for Vy because:

At max throttle, for any given speed, your engine produces constant power (ignoring tedious air density effects). This means some thing or things are steadily gaining energy over time. In this case, flying at a constant speed, your aircraft is NOT gaining kinetic energy (ignoring the tedious air density effects again).

However it is gaining height, which is potential energy. The more power that ends up in this category, the faster you go up. This is "excess power", and it is nice to have some.

As your wing generates lift and drag, you also disturb the air you pass through, handing out kinetic energy to the atmosphere at a steady rate. The more power wasted on the atmosphere, the slower you go up.

For aeroplanes, the power available from the engine depends on airspeed. Power lost to the atmosphere also depends on airspeed. There is a speed, Vy, at which you have the maximum difference between the two, leaving the most "excess power" left over to climb at the fastest rate.

==

The speed for best angle of climb is more complicated.

It also depend on the excess power, ie engine power minus power lost moving the atmosphere around.

You are trying to maximise Rate of Climb divided by TAS.

As we have seen, Rate of Climb is proportional to excess power, so if you plot Excess Power (Engine Power minus power lost to the atmosphere) versus TAS, the point on the graph with the steepest gradient gives you the best ratio of Climb Speed to TAS.

The reason eg Trevor Thom talks about excess thrust is that for a physicist or an engineer, a Power is a Force times a Speed.

Engine Power is Thrust times TAS. Power Required is Drag times TAS. Excess Power is (Thust-Drag) times TAS.

My "max gradient" is Excess Power divided by TAS, ie Thrust - Drag or "Excess Thrust" as Trevor Thom calls it.

Hope that helps.

PS I see that puntosaurus made some of the same points while I was typing, but I've typed too much already, I think...

See How It Flies Is a useful resource. Take a look at chapter 1 to clear up the differences between power and energy But don't get obsessed by this small aspect of your flying, as long as you know that climbing at a lower airspeed than that for optimum rate of climb will cover less ground during the climb thus give you a steeper angle (despite the somewhat poorer rate of climb) For most light aircraft it doesn't make a lot of difference, much less than a few knots of headwind!

Quote:

But what do you mean with:"The more power wasted on the atmosphere, the slower you go up."

I think he is saying the more induced drag you create, the slower you go up. Just another way of putting it......

The more power wasted on the atmosphere, the slower you go up.

Every second, your engine releases some energy from the fuel it burns. Because of the Conservatrion Of Energy, it has to go somewhere. There are three obvious places:

1) Your aircraft can go faster, i.e. get Kinetic energy. But you are climbing at a constant speed, so no energy goes here.

2) Your aircraft can go higher, i.e. acquire Potential Energy. This is what you want.

3) The atmosphere can be moved about, i.e. acquire kinetic energy. You are right, this isn't really "wasted". I should have said it is an inevitable consequence of moving the aircraft through the air (parasite drag) and generating lift (induced drag). Parasite Drag involves a random swirling of air (Kinetic Energy but no momentum), and Induced Drag involves a downwards directed flow of air (Kinetic Energy and momentum).

The energy goes partly to 2, and partly to 3. The more that ends up in 2, the better for your climb.

Incidentally, if you are happy dealing with vector components of forces, there is a more direct explanation for "Excess Thrust".

Consider first a glide approach. Lets say Thrust is actually zero. Drag is still there, and because you are not accelerating (just moving steadily downwards) Lift equals Aircraft Weight (very closely) and Drag is Weight divided by the Lift/drag ratio.

Say LDR=10 so Drag is 10% of aircraft weight. If you descend at 6 degrees, (1 in 10) then 10% of Gravity (ie Weight) is pulling you along and making up the "Thrust Deficit". 10% of say 65kts is about 650 fpm so this makes sense.

If you extend flaps or the prop is windmilling the LDR might be 8, and you need to go down at 1 in 8 to overcome the Thrust deficit and maintain speed. This is about 800 fpm.

In the climb, you need Excess Thrust to overcome a component of Gravity which still depends on your angle of climb, but now works against you.

And I agree with Piper Classique, "See How it Flies" is a wonderful resource!

Last edited by 24Carrot; 12th May 2010 at 09:08.
Reason: Grammar!

dear colleague , can you specify your knowledge level , flying experience, type of a/c flown , to help us to chose the most adequate way of explanation. rgds.

The height in meters that any vehicle can climb is found by using this formula:

ENGINE POWER IN WATTS times POWER LOSS FRACTION AT THE VEHICLE'S SPEED times DISTANCE TO OBSTACLE IN METERS divided by VEHICLE WEIGHT IN KILOGRAMS divided by 9.81 ( Earth GRAVITY FACTOR )

The formula indicates that climb distance will: A. increase in proportion to engine power. B. decrease as efficiency decreases. C. decrease in proportion to vehicle weight. D. decrease as gravity increases.

For and aircraft the efficiency includes:

A Propeller efficiency which is typically about .3

B aircraft lift efficiency this could be about .8 Aircraft lift efficiency normally decreases with speed due to increased drag at higher speeds but lift efficiency also decreases at lower speeds near the stall speed due to increased turbulence.

If the numbers above are used then the total aircraft efficiency would be .3 times .8 which is .24

One horse power is 745.7 watts. One mile is 1609 meters One pound is .45359 kilograms

The calculations indicate that there is a sweet speed for maximum climb distance to an obstacle that is somewhere above stall speed and below higher speeds. That speed is the speed where the aircraft efficiency fraction is highest. This will probably be the lowest speed where you sense that losses due to turbulence and drag are minimum.

This sweet speed should also give the maximum climb distance in a given time because the maximum power from the engine is being transferred into the task of climbing. As far as clearing an obstacle goes, I would error on the side of going slower as long as efficiency seems OK since the aircraft will have more time to expend energy climbing before in hits the obstacle. This will also give you more time to think about how to avoid the obstacle as well as the fact that you will be going more slowly should you hit the obstacle.

My credentials. I took flying lessons many years ago and soloed in a Cessna 152 but I am not a pilot. I was working on electric vehicle charging systems when I came across this discussion while doing vehicle climb rate calculations. You can check out my site at EVfueling.com but I have not spent much time working on the site in the past year or more because I am working on a fueling calculator spreadsheet that will help quantify factors related to vehicle fueling.

As to the OP's question, and applicable at lower altitudes, a climb at Vy (Best Rate) will get you the highest in the least time (greatest FPM), however, you're going to pass over lots of ground while you're doing it. A climb at Vx (Best Angle) will get you higher per given distance across the ground, though you will not be gaining altitude as fast as you would at Vy. Once you have cleared the obstacle, accelerating from Vx to Vy, will give the best climb overall. Flying either type of climb at a differing airspeed will reduce your climb rate to less than the most favourable for that condition.

As for determining climb performance numbers, there's a lot more required to gather actual climb performance data, and correct it for atmospheric and unique aircraft factors than the application of a simple formula. After gathering observed climb performance data (usually a minimum of a dozen or so test climbs per configuration), considerable data reduction is required.

This is in part, because the climb data must be expressed as sea level, standard day values, and you cannot test that way. The non standard altitude affects both power form the engine/propeller, the lift from the wing itself.

The propeller efficiency varies because of many factors, and is not fixed at one value. Generally, a factor around .7 is appropriate, though the application of that factor is complex. An email I received from a McCauley Propeller staff member on this subject recently, reads, in part:

I can say that most parameters the propeller runs about 77% efficient.

Just know which type of climb you need, and fly the numbers in the performance tables for the aircraft for that climb. For sure, having some understanding of the underlying factors is great.

I scanned this thread fast so apols if it has been answered.

'If you climb at vx and you reach 2000' in, let's say, 5nm (at eg 60 mph = 5 mins) while in the mean time you reach in 5nm at vy, 1300' (at eg 120mph= 2.5 mins and 2000' in 3.8 mins but 7.6 miles) why in the hell are going to climb at Vy to gain altitude in the shortest time'

Because in scenario 2 you are going faster, so you hit 2000' quicker ( less time) and you are 2.5 miles and 1.2 minutes ahead of your mate(but if the mountain is at 2000' at 5 miles the point is moot). I think the vx vy numbers you quoted aren't quite that divergent.

Another way of looking at it is if vx=60 and vy=65 and you had a 60 headwind you will clear any obstacle at 60 ( vertical take off) but at 65 you will go up faster ( vertical feet per minute) but will move forward towards any obstacle.

It took me ages to understand the difference. Now you have 17 different views of why it's different . Hope one works out for you.

There seem to be a lot of mathematics based answers here that I'm finding hard to understand. I understood the original question was concerned with the difference between best rate & best angle of climb. My own idea to explain the differences would be that there are two different angles of climb. one is steep but slower, the other is shallow but faster. Over a given time the steep climb will be higher but not so far "down range" the shallow climb will be not so high but further "down range". Placing two 30deg right angle triangles on a table, one with the long "square" side down, the other with the short "square" side down. The hypoteneuse is the time (identical), the horizontal is the airspeed, the vertical is the climbed altitude. Experts please advise if this is bullsh*t.

Over a given time the steep climb will be higher but not so far "down range" the shallow climb will be not so high but further "down range".

Unfortunately not.

Two defined climbs are described as "Best Angle" and "Best Rate".

The "angle" is relative to the surface, so if you stood off to the side of the runway a way back, the airplane flying the "best angle" speed, would appear getting away from earth at an angle less acute relative to the ground, than a plane flying the best rate speed, which would have a more acute, flatter angle. The best angle plane will be moving more slowly across the ground, so combined with the steeper angle, will clear an obstacle that the best rate plane might not.

The "rate" is in feet per minute, for increase in altitude (does not care about distance over the ground covered). So flying the best speed rate speed will get you the greatest altitude in the least time, but you'll cover more ground doing it, so you might hit that same obstacle. So over a given time the shallow angle, faster airspeed, climb (at best rate speed) will get you to the highest altitude.

This is in part because there is more effect of drag at the lower airspeed of the best angle climb. You are flying with less than optimum efficiency, so you will climb at a slower increase in altitude per time (FPM) rate.

These speeds will change with increased altitude. When you get to the absolute ceiling for the plane, they will be the same, and alarmingly close to the faster stall speed way up there.