Why do Vx and Vy meet at the absolute ceiling?

Vx maximises excess thrust and Vy maximises excess power. As altitude increases, power available falls and power required (in IAS terms) rises, so the excess shrinks. At the absolute ceiling power available just touches power required at a single speed, which is then both the best-angle and the best-rate speed, with a rate of climb of zero.

Does Vy increase or decrease with altitude?

For a piston aircraft, Vy as an indicated airspeed decreases with altitude, while its true airspeed increases a little. Vx as an IAS increases slightly. For a jet with thrust independent of speed, Vx stays at the minimum-drag speed and Vy decreases towards it.

What is the relationship between VMP and VMD?

With a parabolic drag polar, the minimum-power speed is the minimum-drag speed divided by the fourth root of 3, about 0.76 times VMD. At VMD parasite and induced drag are equal; at VMP induced drag is three times parasite drag.

What is the difference between absolute and service ceiling?

The absolute ceiling is where the best rate of climb is zero. The service ceiling is a practical limit where the best rate of climb has fallen to 100 ft/min for a piston aircraft.

Vx and Vy: why they meet at the absolute ceiling (climb performance explained)

The sentence everyone learns and nobody explains

Every PPL course teaches two climb speeds. Vx, the best angle of climb, gives the most height per mile of ground in still air, for clearing the trees at the end of a short strip. Vy, the best rate of climb, gives the most height per minute, for getting to cruise altitude. Then comes a sentence most pilots can recite: Vx is always slower than Vy, except at the absolute ceiling, where they are the same.

Most books state it; few show where it comes from. This article shows the mechanism. We write down the forces in a steady climb, turn them into two curves, and watch what altitude does to those curves until the two speeds have nowhere left to go but the same point. There is some algebra and a little calculus; the exams need only the results, which are in bold. An interactive model at the end lets you climb a piston trainer and a light jet all the way to their ceilings.

1. A steady climb in two equations

In a steady, straight climb at a constant airspeed, the aircraft is not accelerating (a simplification we return to at the end of this section), so the forces along the flight path balance. Thrust T pulls forward, drag D pulls back, and a component of weight, W·sin γ, pulls back down the slope, where γ is the climb angle:

T − D − W·sin γ = 0 ⟹ sin γ = (T − D) / W

The climb angle depends on excess thrust, T − D, per unit of weight. The rate of climb is the vertical part of the true airspeed, RC = V·sin γ. Multiply the line above by V:

RC = (T·V − D·V) / W = (P_A − P_R) / WP_A = T·V is the power available, P_R = D·V the power required, V the true airspeed

The rate of climb depends on excess power per unit of weight. Those two lines are the whole article:

Vx is the speed that maximises excess thrust, T − D.
Vy is the speed that maximises excess power, P_A − P_R.

Performance charts and the EASA 032 syllabus usually give this as the climb gradient, height gained per horizontal distance, in per cent: (T − D)/W × 100, since for small angles tan γ ≈ sin γ.

Two simplifications. Strictly, lift in a climb is W·cos γ, a little less than the weight; it only matters through induced drag, and even at the jet's 10° in Exercise 1 (cos γ = 0.985) it changes the angle by about 0.1°, so we treat lift as equal to weight. Second, a climb at constant indicated airspeed (IAS) is not strictly unaccelerated: the true airspeed (TAS) rises as the air thins, and some of the excess power goes into that speed gain. For the trainer this costs less than 1 % of the rate of climb. For the model jet at Vy it costs about 8 % at 10,000 ft, 11 % at 30,000 ft and 17 % at 40,000 ft if the IAS is held constant; following the Vy schedule, whose IAS falls as you climb, the cost is about 2.5 %, 5 % and 10 %. The rates in this article and in the interactive model ignore it, so the jet's best rate in the interactive model is optimistic, more so the higher it climbs. It does not change the absolute ceiling, where the rate of climb is zero anyway.

2. Drag and power curves

A simple drag model is enough to see everything. Drag has a part that grows with speed (parasite drag) and a part that falls with speed (induced drag, the price of making lift):

D = a·V_E² + b / V_E²a = ½ρ₀·S·C_D0, b = 2kW² / (ρ₀·S), V_E the equivalent airspeed.
ρ₀ is sea-level air density, S the wing area, W the weight, C_D0 the zero-lift (parasite) drag coefficient, and k = 1/(π·A·e) the induced drag factor, with A the aspect ratio and e the Oswald efficiency factor.

Written with equivalent airspeed (EAS), this drag curve is the same at every altitude, because both terms depend only on dynamic pressure. For a light aircraft at climb speeds, EAS is practically the IAS; for the model jet high up the difference is a few knots (compressibility), which we ignore. From here on IAS and EAS are the same thing.

Setting the slope of each curve to zero gives the two classic reference speeds:

V_MD, minimum drag: dD/dV_E = 0 gives V_E⁴ = b/a. Here parasite and induced drag are equal, and the lift-to-drag ratio is at its best.
V_MP, minimum power: power required is D·V = D·V_E/√σ, where σ = ρ/ρ₀ is the air density ratio (ρ the density at the aircraft's altitude), and the factor 1/√σ cannot move the minimum, so d(D·V_E)/dV_E = 0 gives V_E⁴ = b/(3a). Here induced drag is three times parasite drag.

Both are fixed EAS values, the same at every altitude.

Rule of thumb

Dividing the two results: V_MP = V_MD / 3^¼ ≈ 0.76 · V_MD. Minimum-power speed is about three quarters of minimum-drag speed, for any aircraft that follows this drag model.

Drag and power required at sea level for our model piston trainer (1,043 kg). Minimum drag at about 68 kt, minimum power at about 52 kt: 0.76 times V_MD, as the algebra says.

3. Where Vx and Vy sit: jet and propeller

The power plant decides the shape of the other curve.

Jet: thrust roughly constant with speed

Vx: if T does not depend on speed, the largest T − D is at the smallest D. So Vx = V_MD. A real turbofan loses some thrust as speed rises at low altitude, which puts Vx slightly below V_MD; Vx = V_MD is the standard simplification.
Vy: power available T·V is a straight line from the origin with slope T. The power required curve D·V has slope D_min, the minimum drag, at V_MD: the line from the origin that just touches it touches it at V_MD. Because T is larger than D_min whenever the aircraft can climb, the gap is still growing at V_MD, so the biggest gap lies above V_MD, well above it when thrust is large.

Propeller: power roughly constant with speed

A piston engine and propeller deliver a power that changes much less with speed, so thrust T = P_A/V falls as speed rises. Low speeds give more thrust, which pulls Vx below V_MD, towards V_MP. With power exactly flat, Vx would even be below V_MP; in our model, where propeller efficiency falls at low speed, it sits between V_MP and V_MD.
If power available were exactly flat, the largest P_A − P_R would be where P_R is smallest: Vy = V_MP. A real propeller becomes more efficient as speed increases, which pulls Vy above V_MP, close to V_MD.

Reading Vx off the rate-of-climb curve

Plot rate of climb against airspeed. Vy is the top of the curve. Since sin γ = RC / V, the steepest climb is where the ratio RC/V is largest: draw a straight line from the origin that just touches the curve. The touching point is Vx. It always lies to the left of the top, so Vx is slower than Vy.

This is in still air. A steady wind does not change the rate of climb, but it changes the angle over the ground: a headwind steepens it, a tailwind flattens it. For the steepest climb over the ground, draw the line from the headwind component on a TAS axis (on an IAS axis, the headwind times √σ) instead of the origin (from minus the tailwind component in a tailwind): the touching point moves a little above Vx in a headwind and a little below it in a tailwind. The flight manual Vx is the still-air value.

On our model trainer at sea level: V_MP ≈ 52 kt, Vx ≈ 56 kt, V_MD ≈ 68 kt, Vy ≈ 71 kt. On the model jet: Vx = V_MD ≈ 144 kt, Vy ≈ 233 kt.

4. What altitude does to the curves

Keep working in EAS. As you climb, the air density ratio σ = ρ/ρ₀ falls, and three things happen:

The drag curve does not move. At a given IAS the drag is the same.
The power required rises. The same IAS means a higher true airspeed, V = V_E/√σ, so P_R = D·V_E/√σ. The whole curve is multiplied by 1/√σ: at 10,000 ft that is about 16 % more. That is on an IAS axis, as in our model. On the TAS axis of most textbook diagrams, every point moves up and to the right by the same factor, along a line through the origin.
The power available falls. A normally aspirated piston engine loses power with density; a jet loses thrust.

The jet, exactly

Vx is the minimum of a drag curve that does not move, so Vx stays at V_MD, the same IAS at every altitude. Its true airspeed rises with altitude, but the indication does not change.

For Vy, write excess power in EAS: (T − D(V_E))·V_E/√σ. The factor 1/√σ does not depend on speed, so it cannot move the maximum. Setting the derivative to zero:

T = D + V_E · dD/dV_E

When thrust is large, this is satisfied far above V_MD, where the drag curve is steep. As thrust falls with altitude, the solution slides back down the curve. At the altitude where the thrust left is just equal to the minimum drag, the only solution is where dD/dV_E = 0, at V_MD. Vy has come down to meet Vx.

The ideal propeller

With power available flat in speed, excess power is P_A(σ) − D·V_E/√σ. The maximum is where D·V_E is smallest, so Vy = V_MP at every altitude. Vx moves instead: the condition for maximum excess thrust becomes

dD/dV_E = −P_A·√σ / V_E²

As P_A·√σ shrinks with altitude, the slope that satisfies it gets flatter, and Vx moves up towards V_MP. At the ceiling, where P_A equals the minimum power required, the two speeds coincide.

A real propeller: both speeds move

Real piston aircraft sit between the two ideal cases, and both speeds move: Vx (IAS) increases a little, Vy (IAS) decreases, and they meet at the absolute ceiling. The Vy trend is the one in the textbooks; the small rise in Vx is what the ideal-propeller result above predicts and what our model shows, and your own flight manual is the authority for your aircraft. Our model trainer, in the International Standard Atmosphere (ISA):

Pressure altitude (ISA)	Vx (KIAS)	Vy (KIAS)	Best rate (ft/min)	TAS at Vy (kt)
Sea level	56	71	820	71
5,000 ft	57	68	580	73
10,000 ft	58	64	360	75
15,000 ft	59	61	140	77
≈ 18,500 ft (absolute ceiling)	60	60	0	80

Look at the last column. Vy as an IAS falls by about 12 kt, but as a TAS it rises by about 8 kt (unrounded model values). Both are true at once, and confusing the two is a classic exam trap.

The models

Trainer: 1,043 kg (W ≈ 10.2 kN), S = 16.2 m², C_D0 = 0.040, A = 7.32, e = 0.75 (k ≈ 0.058). Engine power P = P₀·(1.132σ − 0.132), with P₀ = 134 kW at sea level. Power available P_A = η·P, where the propeller efficiency η = 0.68·[1 − (1 − V/V*)²] rises with true airspeed V to its peak of 0.68 at V* = 58 m/s (about 113 kt TAS) and stays at 0.68 above it.

Jet: 4,700 kg (W ≈ 46.1 kN), S = 20 m², C_D0 = 0.022, A = 8.5, e = 0.80 (k ≈ 0.047). Thrust T = 11 kN · σ^0.8, the same at every speed.

Both: ISA atmosphere, so pressure altitude and density altitude are the same; constant weight; full power; IAS taken as EAS; no compressibility.

5. The ceilings

Absolute ceiling: the altitude where the best rate of climb is zero. Power available just touches power required at one speed, and that one speed is both Vx and Vy.
Service ceiling: a practical limit, the altitude where the best rate of climb has fallen to 100 ft/min for a piston aircraft. Our model trainer reaches it at about 16,000 ft.

You can never actually reach the absolute ceiling. Near the top, the rate of climb falls roughly in proportion to the height still to go, so each further hundred feet takes longer than the one before. In our model, climbing at Vy to 15,000 ft takes about 39 minutes, and to 18,000 ft about 88 minutes. The last 3,000 ft take longer than the first 15,000.

All of this is for one weight, ISA and full power. A heavier aircraft has a larger induced-drag term, so V_MD and V_MP (both proportional to √W), and with them Vx and Vy, rise; the angle and rate of climb fall, and both ceilings are lower. A higher temperature at the same pressure altitude acts like a higher density altitude.

6. Finding Vx and Vy numerically

With a drag polar and a power-available model, a few lines of code find both speeds at any altitude. This is what the interactive model below does for each position of the slider.

PseudocodeVx and Vy at one altitude

function climb_speeds(altitude):
    sigma = density_ratio(altitude)
    best_rate = -infinity;  best_angle = -infinity
    for ias from v_min to v_max step 0.1 kt:
        ve   = ias * 0.5144                # m/s; IAS taken as EAS
        tas  = ve / sqrt(sigma)
        drag = a * ve^2 + b / ve^2         # same curve at every altitude
        excess_power = power_available(tas, sigma) - drag * tas
        if excess_power > best_rate:         best_rate  = excess_power;       vy = ias
        if excess_power / tas > best_angle:  best_angle = excess_power / tas; vx = ias
    return vx, vy, best_rate / weight     # rate of climb in m/s

Climb it yourself

Choose the piston trainer or the light jet and drag the altitude to the ceiling. Watch the green area between power available and power required shrink, the tangent from the origin and the top of the rate-of-climb curve slide together, and the two lines in the last chart meet.

What it means in the cockpit

Use the flight manual figures for your altitude. If the manual gives Vy for several altitudes, the sea-level number is too fast up high.
Vx is close to the stall. In the model trainer, Vx at sea level needs a lift coefficient of about 1.2. The model has no stall of its own; assuming a clean maximum lift coefficient C_Lmax = 1.6 for a light trainer wing (an assumption, not model data), the stall would be at about 49 KIAS, only 7 kt below Vx. Climbing at Vx is for obstacles, and only as long as you need it.
High, hot airfields squeeze the margins. Density altitude shrinks excess power much as altitude does, so the gap between Vx and Vy, and the climb you get at either, is smaller on a summer afternoon in the mountains.
The jet's Vx indication does not change with altitude in this simple model. Real jets climb on an IAS schedule and then, from a crossover altitude, at a constant Mach number. This simple model ignores that, and the aircraft's own performance data is the authority.

Common misconceptions

"Vy increases with altitude"

Its true airspeed does. Its indicated airspeed decreases. The pilot flies the IAS.

"Vx gives the best climb, full stop"

It gives the steepest climb. Vy gives the fastest climb. At any moment after both start from the same point, the Vy aircraft is higher and further along; only at a given distance over the ground is the Vx aircraft higher.

"At the absolute ceiling the aircraft stalls"

Not in itself. In the model it can still fly level at one speed, the one where Vx and Vy meet, with no power to spare for climbing or turning: a turn, or trying to hold altitude as speed decays, means descending or slowing towards the stall. Real jets usually stop well short of this thrust-limited ceiling, because the maximum operating altitude is also limited by the buffet margin (typically 1.3 g to buffet onset), which this model ignores.

"For every aircraft Vy is the minimum-power speed"

Only for an ideal propeller with power available that does not change with speed. For a jet, Vy is well above V_MD.

Exercises

1. A light jet weighs 46.1 kN, has a maximum lift-to-drag ratio (L/D)_max of 15.6 (L the lift, D the drag) and 11.0 kN of thrust at sea level. What is its maximum climb angle?

Minimum drag D = W/(L/D)_max = 46.1/15.6 ≈ 2.96 kN (lift ≈ weight). Then sin γ = (11.0 − 2.96)/46.1 ≈ 0.174, so γ ≈ 10.0°, flown at V_MD.

2. Another trainer, also weighing 10.2 kN, has 70 kW of power available and needs 41 kW at Vy. What is its rate of climb in ft/min?

RC = (70 − 41) kW / 10.2 kN ≈ 2.84 m/s. Times 196.85: ≈ 560 ft/min.

3. An aircraft's minimum-drag speed is 80 KIAS. Estimate its minimum-power speed.

V_MP ≈ 0.76 × 80 ≈ 61 KIAS, from V_MP = V_MD/3^¼.

4. Why is a jet's Vx the same IAS at every altitude in the simple model?

With thrust independent of speed, Vx is the minimum of the drag curve, and in terms of IAS (EAS) the drag curve is the same at every altitude. Only the thrust line moves down.

5. As a piston aircraft climbs, what happens to Vy as an IAS and as a TAS?

The IAS decreases and the TAS increases. In our model: 71 KIAS / 71 KTAS at sea level, 64 KIAS / 75 KTAS at 10,000 ft.

6. What is the rate of climb at the absolute ceiling, and how many speeds can the aircraft fly level at there?

The rate of climb is zero, and there is only one speed at full power, where Vx and Vy coincide; at any other speed it must descend.

Practise for the ATPL and PPL exams

Free exam-style questions with worked answers, chapter by chapter.

Open the questions ›

FlightDesk Aero articles are for training and reference. The aircraft in this article are simplified models built to show the mechanism; their numbers are not data for any real aircraft. Always fly the speeds in your aircraft's flight manual (POH/AFM). Sources: Oxford ATPL 032 Performance and 081 Principles of Flight (reference); AOPA Flight Training, "No dumb questions" (Nov 1998); standard climb-performance theory (drag polar, ISA atmosphere).

Altitud de presión (ISA)	Vx (KIAS)	Vy (KIAS)	Mejor régimen (ft/min)	TAS a Vy (kt)
Nivel del mar	56	71	820	71
5.000 ft	57	68	580	73
10.000 ft	58	64	360	75
15.000 ft	59	61	140	77
≈ 18.500 ft (techo absoluto)	60	60	0	80

Altitudine-presiune (ISA)	Vx (KIAS)	Vy (KIAS)	Viteză verticală max. (ft/min)	TAS la Vy (kt)
Nivelul mării	56	71	820	71
5.000 ft	57	68	580	73
10.000 ft	58	64	360	75
15.000 ft	59	61	140	77
≈ 18.500 ft (plafon absolut)	60	60	0	80

The sentence everyone learns and nobody explains

1. A steady climb in two equations

2. Drag and power curves

3. Where Vx and Vy sit: jet and propeller

Jet: thrust roughly constant with speed

Propeller: power roughly constant with speed

4. What altitude does to the curves

The jet, exactly

The ideal propeller

A real propeller: both speeds move

5. The ceilings

6. Finding Vx and Vy numerically

Climb it yourself

What it means in the cockpit

Common misconceptions

"Vy increases with altitude"

"Vx gives the best climb, full stop"

"At the absolute ceiling the aircraft stalls"

"For every aircraft Vy is the minimum-power speed"

Exercises

Practise for the ATPL and PPL exams

La frase que todos aprenden y nadie explica

1. Un ascenso estabilizado en dos ecuaciones

2. Curvas de resistencia y de potencia

3. Dónde están Vx y Vy: reactor y hélice

Reactor: empuje casi constante con la velocidad

Hélice: potencia casi constante con la velocidad

4. Qué hace la altitud con las curvas

El reactor, exactamente

La hélice ideal

Una hélice real: se mueven las dos

5. Los techos

6. Cómo encontrar Vx y Vy numéricamente

Súbelo tú

Qué significa en la cabina

Confusiones frecuentes

«Vy aumenta con la altitud»

«Vx da el mejor ascenso, y punto»

«En el techo absoluto el avión entra en pérdida»

«En cualquier avión, Vy es la velocidad de potencia mínima»

Ejercicios

Practica para los exámenes ATPL y PPL

Fraza pe care o învață toți și n-o explică nimeni

1. O urcare stabilizată în două ecuații

2. Curbele de rezistență și de putere

3. Unde stau Vx și Vy: reacție și elice

Reacție: tracțiune aproape constantă cu viteza

Elice: putere aproape constantă cu viteza

4. Ce face altitudinea cu aceste curbe

Avionul cu reacție, exact

Elicea ideală

O elice reală: se mișcă amândouă

5. Plafoanele

6. Cum găsești Vx și Vy numeric

Urcă tu

Ce înseamnă asta în cabină

Confuzii frecvente

„Vy crește cu altitudinea”

„Vx dă cea mai bună urcare, și gata”

„La plafonul absolut avionul intră în angajare”

„La orice avion, Vy este viteza de putere minimă”

Exerciții

Exersează pentru examenele ATPL și PPL