Why is the initial short-field climb below Vₓ in a Cessna 172?

Question

Short-field takeoff procedure for a Cessna 172R is

1. Wing Flaps – 10°.
2. Brakes – APPLY.
3. Throttle – FULL OPEN.
4. Mixture – RICH …
5. Brakes – RELEASE.
6. Elevator Control – SLIGHTLY TAIL LOW.
7. Climb Speed – 56 KIAS (until all obstacles are cleared then $V_y$).
8. Wing Flaps – RETRACT slowly after reaching 60 KIAS.

However, best angle-of-climb or $V_x$ at is 62-67 KIAS depending on elevation. Why is the first stage of the climb slower than $V_x$, and if the slower speed gives better performance, then why isn’t that the published figure?

Answer 1

The published Vx is in the clean condition (no flaps): 62.

With the flaps down, for a short-field takeoff, Vx is 56.

This is the same reason why the stall speed (bottom of the white arc) is lower with flaps than without (bottom of the green arc)

Answer 2

$V_X$ published in the manual is $V_X$ for flaps up. $V_{X_{F10}}$ will be slower than $V_{X_{UP}}$. Most likely, the speed in the procedure is, or is very close to $V_{X_{F10}}$.

$V_X$ (speed for angle of best climb) occurs where the excess thrust ($T$ - $D$) is the highest. Refer to the graph below:

$D$ is Induced drag ($D_I$, varies with $\frac{1}{v^2}$) + Parasite drag ($D_P$, varies with $v^2$). By extending the flaps, you are increasing only $D_P$ and the graph describing $D$ ($D_{tot}$ in the graph) moves upwards to the left. This means that the point for $V_{MD}$ moves to the left. Since $V_X$ is found at max $T - D$ it varies with thrust too (which will vary differently, depending on the type of propulsion system), and is not automatically found at $V_{MD}$, but the two speeds generally move together in the same direction.

There's also another, albeit marignal, effect (that i did not put in the graph) to take into consideration for the first 10-20 ft, and that's the ground effect. In ground effect, the aircrafts $D_I$ will be lower, further moving the $D$ graph to the left, and downwards, resulting in a lower $V_{MD}$, and therefore generally also lower $V_X$

Answer 3

Small GA POHs can be fairly sloppy (and your checklist seems even looser, what is your Vr? Why are you going rich after full throttle?).

POH's often will fail to properly define things. For example is a short field(aka maximum performance) takeoff the shortest ground roll, or the shortest distance to clear a 50ft obstacle? That difference needs to be known before you can get a meaningful answer to your question.

Vx (62) is best steady-state climb angle at standard sea level, like if you have a large hill to climb over, it increases and approaches best glide speed at your absolute ceiling.

However, a takeoff is not a steady state condition, it is a dynamic combination of factors. The best 50ft obstacle distance is the best angle measured from the beggining of the runway(brake release point), not the angle measured from the end of rotation. Depending on density altitude, gross weight, and especially headwind; 56kIAS with flaps-10 may shorten ground roll and start the climb earlier more than compensating for a lower steady-state angle of climb.

This isn't even adding the effects of a zoom climb. (Although the POH procedure would be based on such a maneuver.) Increase speed on the ground or in ground effect, then continue rotation to a higher pitch and convert speed into altitude pitching down slightly as you get to Vx. Actual advantage will vary based on many factors.