# How does gross weight affect the Minimum Controllable Airspeed (Vmc)?

I am about to get my MEL added to my CFI, and I cannot get my head around the multiple explanations for how gross weight affects Vmc.

The accepted answer is that The higher the gross weight, the lower is the Vmc. I have been proffered two contradictory explanations, from both internet articles, manufacturer documents, and other CFIs, both of which seem to me (I have an MSME masters degree in Aero) to be flawed.

1. That because the aircraft is slightly banked into the direction of the good engine, the horizontal component of Lift (HCL) opposes the yaw from the rudder, thus allowing the aircraft to slow to a lower airspeed before full rudder authority is required to counteract the yaw from the operating engine. This is wrong, because if you're in a bank, the HCL (by definition, it is HORIZONTAL!), is no longer lined up with the lateral axis of the aircraft. It is misaligned by exactly the bank angle, so, although there is a component of the HCL opposing the yaw force from the rudder, there is an exactly equal and compensating component of the Vertical Component of Lift (VCL) that will be augmenting the Yaw force. This argument is bogus.

I mean if you think about it simpler, the Lift vector is always, ALWAYS, perpendicular to the wings, so that any effect of Lift along the lateral axis (parallel with yaw forces), must necessarily be zero.

The fact that is not aligned with the earth horizon is irrelevant.

1. The other rationale I have been presented with is that as the gross weight increases, the resistance to motion (Engineers would call this rotational inertia) increases, and makes the aircraft more stable. This is true, but it is an argument about dynamic stability, not about static stability. In other words, this affects the aircrafts resistance to changes in yaw/sideslip angle. Vmc is about static stability, i.e., at what airspeed does the aircraft, in a static, (unchanging) zero sideslip angle, require full rudder defection to hold the aircraft (STATICALLY) at that zero sideslip angle against all the yaw forces being produced by all the factors resulting from Asymmetric thrust?

So this argument or rationale also seems to be incorrect to me.

Where is the flaw in this reasoning?

By the way, the only effect of Gross weight on Vmc that I can think of is the obvious one, The higher the gross weight, the greater the Angle of Attack required to hold 1 G (Level flight), and obviously, both P-Factor and adverse yaw increases with AOA. So the higher the Gross weight, the greater the P-Factor, and the greater the yaw-induced aerodynamic effects of the asymmetric thrust. So, if this logic is correct, Higher Gross weight means Higher Vmc, not Lower.

I think your objection to #2 is exactly correct, and your #1 is where the misunderstanding may lie.

The point of VMCA is essentially to maintain a constant heading, and when you have no more rudder authority to make that happen, you're there -- and any slower, with slightly less rudder authority, your heading starts to change in the direction of the dead engine. With no bank, it's all rudder; the lift from the wings doesn't tend to move the nose in any direction. With bank, there is now a component of the lift from the wings that does tend to affect heading (which is the dynamic of banking when you turn -- the heading change comes from the lift created by the wings, not from the lift generated by the rudder).

If you are banking into the good engine, that component of lift tends to change your heading toward the good engine and away from the failed engine. Since the asymmetric thrust is working to change your heading toward the failed engine while you hold rudder to oppose that change, you now have the component of lift from the wings that is assisting your rudder.

If you consider a case of symmetric thrust, you can hold a heading with right wing down and left rudder or vice versa; the lift from the wings is working to change your heading to the right, and your rudder is counteracting that drift. At some point, with enough bank, you run out of rudder authority & the airplane will turn, although pretty badly uncoordinated. With an engine out, you CAN fly straight ahead holding bank into the dead engine, you just need a lot of rudder authority (i.e. lots of speed above VMCA) to hold your heading -- the rudder is fighting both the asymmetric thrust and the component of the wings' lift. And doing that, you'll run out of rudder authority a lot sooner (i.e. at a higher airspeed). If you switch to bank into the good engine, the component of lift from the wings is now working with you, and that's the usual case for computing and demonstrating VMCA.

I suspect that the confusion on your point #1 comes by conflating horizontal (earth reference) with horizontal (aircraft reference). Hopefully the explanation above separates things out?

And, to bring things back to gross weight, the higher the weight, the greater the lift from the wings, so a Sin(5 degrees) (I think that's right -- LONG time since I took Trig!) component of "more lift" is a greater force to resist heading change than that same component of "less lift" with a lighter aircraft. The other forces involved in the balance, the asymmetric thrust and the rudder force, are independent of aircraft weight.

Best wishes for your MEL checkride!

• But the vertical component of Lift is acting in the opposite direction. i.e., Say the vertical component of Lift is LCos(5 deg). Then the component of that which is acting along the aircraft lateral axis (parallel to the Yaw force) is LCos(5deg)sin(5deg), in the SAME direction as the Yaw force, so it is augmenting the effect of yaw from the Rudder. The exacerbating effect of the HCL is Lsin(5deg), and the component of that which is aligned with the Yaw force is Lsin(5deg) cos(5deg), in exactly the opposite direction. They cancel each other out. – Charles Bretana Dec 15 '17 at 19:13
• By the way, as I understand it, the reason we are in a bank is because although we need the rudder to attain zero sideslip (to minimize drag and maximize excess thrust), this generates a sideways yaw force, which, uncorrected, will cause the aircraft to turn. We put the aircraft in a bank to counteract this turning force and keep a constant heading. – Charles Bretana Dec 15 '17 at 19:37
• @CharlesBretana Let's say your left engine is dead. Aircraft wants to yaw left. Rudder is displaced right, pushes tail left, and nose right. You're banking right (into the good engine). The lift from the wings is: all vertical (aircraft reference), mostly vertical(earth reference), some right (earth reference). Which tends to track the nose to the right -- assisting the rudder. The rudder is "pushing (the tail) left" while the wings are "pushing (the nose) right" but they're both pushing the heading in the same direction (right). – Ralph J Dec 15 '17 at 20:37
• Also, NO component of lift from the wings ever generates ANY yaw. Those two components are always perpendicular. The lift from the wings can track the nose, but it doesn't generate yaw. When you talk about a component of lift acting parallel to the yaw force, that's not quite accurate. Your "wing-aligned component of VCL" isn't right -- the LIFT vector has ZERO wing-aligned component. It has a horizontal component in the earth-frame-of-reference, but not in the aircraft frame of reference. Your red & purple arrows will cancel each other out; the blue HCL arrow remains & tracks the nose. – Ralph J Dec 15 '17 at 20:48
• Right - the lift vector CANNOT generate Yaw. That's my whole point here. No matter how much bank you have, ALL Lift is perpendicular to the Yaw vector. What I am describing is just Math to address the argument that the Horizontal component of Lift (HCL) somehow augments and exacerbates the yaw from the rudder, which is the essence of this argument. It can't, because if you want to actually look at it this way, then you also have to look at the effect of the Vertical component of Lift, which (because of the argument you make), must exactly oppose and compensate, for the effect of the HCL. – Charles Bretana Dec 15 '17 at 22:45

The flaw in the reasoning under 1. is that you should not show horizontal and vertical components of lift, but of gravity. You're considering the aircraft reference frame, and should look at gravity alignment with the aircraft axes. This is proportional to sin $\Phi$ and to mass.

What you were doing was breaking the lift vector into earth axis components, then re-breaking the resulting components into aircraft axis components where they already were.

Well, let’s consider the problem in detail.

Assumptions:

the aircraft is a light twin under 6,000lbs GTOW powered by reciprocating engines.

Engines and propellers are of a tractor configuration, mounted on pylons forward of the wing leading edges. Engines are not geared to counter rotate and turn clockwise from the pilot’s perspective; left engine is the critical engine.

The center of lift is aft of the center of gravity for the entire approved CG Range.

Factors affecting Vmc is the position of the CG as this affects the length of the moment arm for the rudder to counteract the vertical moment caused by the asymmetric thrust loading as a consequence of a critical engine failure. Aft most CG is the worst case.

Asymmetric thrust moment will be a sum of the moments caused by the position of the line of thrust (engine crankshaft) relative to the lateral position of the aircraft CG (airplane centerline) and the thrust aysymmetry caused by P Factor over the propeller disc. Worst case is right at takeoff power at critical AOA.

The horizontal component of lift HCL is used to counteract the lateral drift resulting from the yawing force of the rudder. Regulations and good operating practices limit the amount of bank used to generate this HCL to 5° in order not to diminish the climbing characteristics at low speed with a OEI condition.

Since the airplane will reach the critical AOA at a higher indicated airspeed when at GTOW than at a lower gross weight, it will stall at faster airspeeds than a lightly loaded aircraft would. But since the force and the corresponding counter moment that the rudder can generate are greater for a higher airspeed than a lower one, the aircraft may stall at or below the minimum control speed in this case. At a lower loading the aircraft will stall at a slower airspeed, but the rudder will have a diminished effectiveness here. This may result in a case where Vmc may be reached at or above the stall speed for the aircraft. The wings as well are generating greater lift for an AOA at higher speeds, so the HCL can more effectively counter the rudder loads at higher airspeeds than at lower ones without exceeding the 5° bank limit for this.

In truth, there is no one set Minimum Control Speed; it varies based upon a number of factors such as aircraft type and configuration, density altitude, gross weight, etc. The value of Red Line is based upon certain configurations during flight testing. In practice, minimum control speed will be identified when you encounter any of these three conditions with OEI and the propeller windmilling:

• Loss of directional (yaw) control.
• Maximum rudder deflection in the direction of the operating engine.
• Buffet just prior to aerodynamic stall.

Recognizing this condition and being able to articulate that to a student pilot next to you is what an examiner is going to want you to demonstrate on your checkride and is probably going to be the limit of the knowledge they expect about Vmc.

• Gravity has no aerodynamic effect on an aircraft. Vmc is an aerodynamic effect. Gravity has no aerodynamic effect on an aircraft. Only aerodynamic effects can affect Vmc. – Charles Bretana Dec 17 '17 at 3:02
• Secondly, the rudder is not the only aerodynamic surface creating lateral lift. The fuselage can create lateral lift. That's why you need a right bank to counteract the left turn induced by Left sideslip when landing in a crab in a right crosswind. Left rudder does create right lift, but it puts the fuselage into a left sideslip (to line up with the runway), and the left sideslip creates a larger lift to the left. With left engine out, right rudder only brings the fuselage back to zero sideslip, - the only lateral lift is from the rudder, to the left. So the right bank counteracts that. – Charles Bretana Dec 17 '17 at 3:10
• And my point in breaking the lift into components twice is just to show that this technique is indeed, pointless. The Lift is always perpendicular to the lateral axis of the aircraft, so Breaking it up into Horizontal and vertical components (relative to the horizon), is meaningless, since these are not aligned with the aircraft. And breaking those up into aircraft aligned components, just reverses the process and shows that there is zero effect along the lateral axis from any amount of lift, no matter the gross weight. – Charles Bretana Dec 17 '17 at 3:16
• Charles, you're getting tangled up here with your information and it's causing you to miss the real facts. The HCL is necessary in this problem as it is the force causing the aircraft to 1) turn or 2) counter the lateral lift force from the vertical fin and rudder during to prevent lateral drift of the aircraft. You use right rudder in the event of a left engine failure to counter the torque generated by the asymmetrical thrust (do a sum of moments about the CG with an OEI and this becomes clear). – Carlo Felicione Dec 17 '17 at 9:49
• Carlo,I'm sorry, but I don't believe I am. The HCL, as defined, is the component of Lift parallel with the Horizon, due to the bank, and yes, it, alone, would cause the aircraft to turn to the right. I fully understand that. But The aircraft is NOT turning right, so something else must be causing it to remain on a constant heading. That something is the leftwards Lift generated by the right rudder. That's why we establish the right bank - to stop the left turn that the rudder would induce. But what does this have to do with Vmc??!! – Charles Bretana Dec 17 '17 at 16:22

It's fun to listen to engineers talk about flying airplanes. I know this is a little late for the answer but here we go. It's not the lift, It's the weight. A plane in flight balances at the center of lift, not the center of gravity. As the CG is normally forward of the CL the bank causes the nose to yaw toward Gravity. The more weight vectored toward the live engine with bank the less the rudder has to do. The bank angle is limited because of other considerations. 90* of bank would vector all the weight to oppose yaw. This is the same reason that excess thrust, not more lift, is the force that causes the constant speed climb. Anything that causes less lift to have to be generated by the vertical tail will cause the speed needed to produce that lift to be lower. Changing the weight vector or increasing the vertical tail moment by moving the CG forward will decrease the total lift required to counter the yaw.

• Just as a note, answering older questions is fine on Stack Exchange, as long as you're able to add something that hasn't already been said in other answers and you make sure to actually answer the question that is being asked rather than simply providing commentary on existing answers. – a CVn Feb 7 '18 at 20:12
• @user28842, Aircraft balance aerodynamically about the Center of Lift, they "rotate", inertially, about the CG. Second, I believe that when a body is airborne, the force of gravity is a fictitious force, only necessary in calculations because we do our calculations in an accelerated frame of reference. The only forces that need be considered for this issue are the aerodynamic forces, and the asymmetric thrust of the one good engine. – Charles Bretana Feb 8 '18 at 1:59
• @CharlesBretana “Gravity is a fictitious force” in flight? Some force needs to balance Lift, since the aircraft is often maintaining an altitude rather than accellerating upward at (Lift/Mass) ft/sec/sec for the entire flight... – Ralph J Feb 8 '18 at 4:05
• @RalphJ, Actually no, its is just like Centrifugal "force", in that it is a fiction, a mathematical construct only required to make the calculations correct when using an accelerated frame of reference. Imagine doing the same calculations in outer space, with no gravity, but inside of a giant box that was accelerating under one "G" of thrust. You would be of course have to include one "G" of gravity to get the answer to be correct relative to the frame of reference of the box. – Charles Bretana Feb 8 '18 at 12:09
• This is same idea as Einstein's famous elevator experiment, thegreatcoursesdaily.com/einsteins-experimental-elevator where he compared being in outer space, in free fall, and being in a free-falling elevator... They are not similar, they are the same, because they are both in a zero-G or non-accelerating frame of reference, – Charles Bretana Feb 8 '18 at 12:11