Class A Aircraft - Take-off — Page 383, Lesson 471

I want to walk you through the relationship between V1, VMCG, and VMBE — because this is where the take-off speeds start to constrain each other, and it's a classic exam area. First, let me set the scene. We've already met VMCG and VMBE separately. VMCG is the minimum control speed on the ground — the lowest speed at which, after an engine fails, you can still maintain directional control using rudder alone. VMBE is the maximum brake energy speed — the highest speed from which you can bring the aeroplane to a stop without overheating the brakes beyond their energy limit. Now, the key rule: V1 must not be less than VMCG. Why? Because if an engine fails below VMCG, the aeroplane is uncontrollable — you simply cannot keep it straight on the runway. And the very definition of V1 is that the take-off can be continued following engine failure. So if V1 were set below VMCG, you'd be committing to a continued take-off at a speed where you can't control the aircraft. That's not allowed. The second rule: V1 must not be greater than VMBE. Again, think about what V1 means — at V1, the aeroplane must be able to either stop or continue the take-off. But above VMBE, it is impossible to bring the aeroplane safely to a stop, because the brakes simply cannot absorb that much energy. So V1 is boxed in: it can't go below VMCG, and it can't go above VMBE. Now let me walk you through a scenario that shows why this matters. Suppose that due to high density — that's high air density, which affects control surface effectiveness — the value of VMCG comes out higher than the idealized V1. In that case, take-off is prohibited. You can't use that V1 because it violates the rule. But the problem is solvable. The chosen V1 can simply be increased until it is equal to or more than VMCG. That brings you back inside the rules. However — and this is the important consequence — notice what happens to the distances. The accelerate-stop distance increases. The take-off distance decreases. And more importantly, the total field length required increases. Let me make sure you understand why. Accelerate-stop distance is the distance to accelerate to V1, then lose an engine and stop. If you raise V1, you're committing to a higher speed before you decide to stop, so you need more runway to stop — that's why accelerate-stop distance goes up. Take-off distance is the distance to accelerate to V1, lose an engine, and continue to take-off. A higher V1 means you're closer to rotation speed when the engine fails, so the continued take-off is shorter — that's why take-off distance goes down. But the total field length required is the governing figure — it's the longer of the two, and it increases. So the runway must be at least as long as this new total field length. So long as the runway is as long as the total field required, then moving V1 to this point is not a problem. But if the runway is short, you may have a real constraint. Let me also flag the factors that change VMBE — temperature, mass, slope, and wind. Carefully examine each of these, because they all shift where VMBE sits. And remember, CAP 698 is for use in the exam — so if there are any questions relating to VMBE, a lot of the information you need is already in front of you in that document. Let me show you this graphically. Here's the picture. You can see VMCG on the left, VMBE on the right, and V1 sitting between them. The shaded region between VMCG and VMBE is where V1 is allowed to live. If V1 tries to go below VMCG, you lose control on engine failure. If it tries to go above VMBE, you can't stop. So the balanced or ideal V1 sits somewhere in that window — and if pressure from VMCG or VMBE forces it to move, you must re-check the distances, because the total field length required will change. So the takeaway: V1 is not a free choice. It's bounded below by VMCG and above by VMBE, and moving it within that window has real consequences for accelerate-stop distance, take-off distance, and total field length required.