# Why does Vx have a relationship with thrust but Vy has a relationship with power?

https://www.boldmethod.com/learn-to-fly/performance/vx-vy-altitude-and-where-they-meet/

After I watch this article, now I understand the relationship between altitude and the rate of climbs.

But I wonder why Vx has to do with thrust and Vy has to do with power.

Could you explain them for me?

For the best climb speed, you need to maximize your vertical speed $$\text{v}_z$$.

For the best climb angle, you need to maximize the ratio $$\frac{\text{v}_z}{\text{v}}$$ of vertical speed $$\text{v}_z$$ and flight speed v.

And, as Robert correctly notes, thrust is power divided by speed. $$\text{T}=\frac{\text{P}}{\text{v}}$$. Do you see a parallel?

Please see here for what needs to be considered to calculate climb speeds at various altitudes. To simplify things, we can model the airplane by using the quadratic drag polar and assume that thrust depends on air density and on flight speed to the power of a constant exponent n$$_v$$ which normally is -1 for piston airplanes. This gives us three equations to model lift L, drag D and thrust T. Weight is given and considered constant by neglecting fuel consumption.

$$L = m \cdot g$$

$$D = \frac{\rho}{2}\cdot v^2 \cdot \left(c_{D0} + \frac{c_L^2}{S\cdot AR\cdot\pi}\right)$$

$$T = T_0 \cdot \rho \cdot v^{n_v}$$

How thrust changes over altitude is explained here (besides much else). The equation is valid for a wide subsonic speed range but does not cover static thrust. Now we can express climb speed as the excess power left over after subtracting drag from thrust, times flight speed:

$$v_z = v\cdot\frac{T-D}{m\cdot g} = \frac{v}{m\cdot g}\cdot \left(T_0 \cdot \rho \cdot v^{n_v} - \frac{\rho}{2}\cdot v^2 \cdot \left(c_{D0} + \frac{c_L^2}{S\cdot AR\cdot\pi}\right)\right)$$

When we now express the lift coefficient as a factor of speed, the equation is ready for derivation with respect to speed. We do the same for the equation for the climb angle. Set both results to zero to find the conditions for the highest climb speed and climb angle.

$$c_L = \frac{m\cdot g}{\frac{\rho}{2}\cdot S\cdot v^2}$$

$$\Rightarrow\:v_z = \frac{v\cdot T_0 \cdot \rho \cdot v^{n_v}}{m\cdot g} - \frac{\rho\cdot S\cdot v^3 \cdot c_{D0}}{2\cdot m\cdot g} - \frac{2\cdot m\cdot g}{\rho\cdot S\cdot v\cdot AR\cdot\pi}$$

and

$$\frac{v_z}{v} = \gamma = \frac{T_0 \cdot \rho \cdot v^{n_v}}{m\cdot g} - \frac{\rho\cdot S\cdot v^2 \cdot c_{D0}}{2\cdot m\cdot g} - \frac{2\cdot m\cdot g}{\rho\cdot S\cdot v^2\cdot AR\cdot\pi}$$

Or you put them in a spreadsheet and plot them over speed. Either way. But what you can see already is that in the climb speed equation we have power (thrust times speed) on the right side whereas in the climb angle equation only thrust is left on the right side.

Having read the article you link to, I am not very happy with the explanations given there. The power curve is supposed to change as you climb - Nonsense! What changes is your true air speed, but the plot sticks with the same true air speed range as altitude increases. All it shows are ever smaller sections of the power curve, which creates the illusion of change where no change is. Plot it over IAS and see for yourself.

Nomenclature:
$$c_L \:\:\:$$ lift coefficient
$$c_{D0} \:\:$$ zero-lift drag coefficient
$$n_v \:\:\:$$ thrust exponent, as in $$T \sim v^{n_v}$$
$$D \:\:\:\:$$ drag
$$S \:\:\:\:$$ wing area
$$T \:\:\:\:$$ thrust
$$T_0 \:\:\:$$ density-specific thrust at a given speed and altitude
$$m \;\:\:\:$$ mass
$$g \:\:\:\:\:$$ gravitational acceleration
$$v \:\:\:\:\:$$ flight speed (as true air speed)
$$\gamma \:\:\:\:\:$$ flight path angle
$$\rho \:\:\:\:\:$$ air density
$$\pi \:\:\:\:\:$$ 3.14159$$\dots$$
$$AR \:\:$$ aspect ratio of the wing

The angle a plane can climb at is entirely dependent on excess thrust. With props, maximum thrust can be at a lower airspeed than Vbg, so even if the plane is not at its optimal wing AOA (more drag), the maximum line between drag and thrust is where you get your best climb angle, Vx.

That's the easy part. How can a plane climb faster when it is climbing at a lower angle? By going faster!

The formula for Power is Force (thrust) x Velocity

When you see this graph, note that the maximum distance between Power and Drag is Vy. This plane will fly higher and farther for a given amount of fuel than at Vx.

Vx is conceptually pretty simple, a climb is very similar to a car going uphill. The steeper the hill the more excess thrust is needed regardless of speed. The only real difference is that a car only has parasite drag and can down-shift to increase thrust at lower speeds, so assuming infinite traction the VX of a car approaches 0 speed. In a plane there is a need to maintain lift equal to aircraft weight. The resulting induced drag and the efficiency of the prop at different airspeeds(The prop also has its own angle of attack) all play together so that maximum excess thrust happens at a particular speed; usually close to minimum total drag depending on the prop selection and engine torque curve.

Now Vy is conceptually a little less clean and doesn't have the direct car analogy(a car's Vy is also its Vx). But it is still a matter of going uphill. If the prop were optimized only for rate of climb then Vy would occur very close to but never below the point of maximum lift/drag ratio for basically the same reasons that best glide distance(angle) is at best L/D and best cruise range is always above best L/D.(Sorry, To be continued something just came up.)