# Would a stationary WWII fighter fall on its nose with full throttle and brakes on?

Would a stationary WWII fighter fall on its nose with full throttle and brakes on?

It's a general question, but I am interested in the Republic P-47.

• When you say "full throttle", I assume you're gunning it from a dead stop? Otherwise, a stationary object just sits there... because it's stationary. – Omegacron Apr 7 '17 at 12:10
• The aircraft is stationary with engine running and brakes on and tail resting on ground. I would then throttle up to full power. – Invariant Apr 7 '17 at 12:54

(Wikimedia; arrows and CG icon are my additions) P-47 side drawing.

The thrust has a bigger moment arm around the tires contact points compared to the weight. It won't nose over below a certain RPM (it will be specified in the training). Below that RPM:

• The weight will counter the thrust
• The propeller downwash will produce downward force on the horizontal stabilizer
• Applied brakes will resist the rotation of the wheels (airframe rotating around wheels)

From a training video for the comparable big-engined F6F Hellcat, the narrator says:

You can't gun this engine to full takeoff RPM and manifold pressure while holding the plane with the brakes, if you exceed 2,000 RPM the tail will lift and you risk nosing over.

Answer is yes it can happen.

Note: for comparison, the max RPM is 2,700 for the Hellcat.

• Okay, does anyone know any introductory text available on the Internet that describes the pitch moment effect from propeller airflow, maybe with some example pitch coefficient curves? – Invariant Apr 6 '17 at 23:49
• The problem is that the movement arm would be based on the wheels, not the CG, so there is definitely a meaningful movement arm. Up elevator would probably be enough to counteract that, but (maybe) neutral or (especially) down elevator would probably get you in trouble. – Eugene Styer Apr 7 '17 at 0:39
• "Applied brakes will resist the rotation of the wheels (airframe rotating around wheels)" - But they are wheels. The whole structure is free to rotate around this 'locked' axle, so I don't see how this has any stabilising effect. Nor does this change when you throttle up (unless tire friction is low enough that you start to slide, which would actually help prevent a dive) – Baldrickk Apr 7 '17 at 8:49
• Now I'm watching an F6F instructional video at work. I'm.. pretty sure that doesn't relate to my job... – enderland Apr 7 '17 at 18:32
• Pity the poor sod who forgot to take his brakes off and worked out where the biting point is. – Richard Apr 8 '17 at 18:57

This was also an issue when doing maintenance. For engine ground running at full power, the tail of the aircraft was weighted down to stop the aircraft tipping over the front wheels.

You can see the "saddle bag" and weights hung over the fuselage in front of the tailplane in this still from a video of a WWII Spitfire ground run:

• Where exactly are the weights? It looks like the plane is tied down to the ground in those pictures. – Andrew Medico Apr 7 '17 at 12:09
• @AndrewMedico, eh? The ground isn't heavy enough? – Wossname Apr 7 '17 at 13:03
• @Wossname clearly, you're underestimating the power of an angry Merlin engine being restrained for emergency surgery. – Jules Apr 7 '17 at 18:30
• @AndrewMedico the weights are the two lumps of lead hanging either side of the fuselage. Lead is pretty dense stuff! The plane certainly isn't "tied down to the ground" - I've seen this done "live" several times. – alephzero Apr 7 '17 at 19:52
• Note the full elevator deflection. The pilot has the stick buried in his gut to make sure that the horizontal stab/elevator are generating significant down-force. – Bob Jarvis - Reinstate Monica Apr 9 '17 at 2:04

# Depends on the elevator input

As long as the stick is held back, as it should be for any kind of runup, then the elevator would be applying plenty of downward force.

Now, without the elevator input to cancel it, the pitchover torque might nose over an airplane, and it might not take one of those big WWII engines to do it.

There is a force directed aft at the interface where the (brake-locked) wheels meet the ground that is equal to the force forward along the propellor shaft. This force is multiplied by the distance between the extension of the prop shaft and the ground between the wheels to create the torque.

It's the same torque, but with opposite magnitude, as that which produces the hotrod wheelie maneuver. (And the torque is there with or without the wheelie, which is why FWD cars don't make good drag racers.)

It will be worse for an airplane with long gear legs that sits well off the ground.

If the static thrust or the elevator moment is high enough, it could happen. I've always wondered what would happen if I didn't hold the stick back on runup.

Update: Addressing comments ... well, it might take a little unfavorable wind too, certainly airplanes have nosed over, see below, on (perhaps downwind) runups, but note that on the takeoff roll in a tailwheel airplane, the tail comes up quite early on, and that's with neutral stick inputs, so one can only imagine that it could happen pretty darn fast with any reckless amount of forward stick.

Apparently, even nose wheel airplanes can do it, I believe this is a Cessna on the runup pad at KSQL:

• Could you estimate (or measure?) the airspeed at the tail when stopped but with the throttle at taxiing speed and at full throttle ? – Criggie Apr 7 '17 at 21:28
• Is really the jet produced by the propellor so strong for this to be true? – Vladimir F Apr 8 '17 at 11:48