Why does V1 increase with weight? (B738 AFM, 0-2000 feet elevation, up to 27°C) Example of V1 increasing with weight.



This valid question came up the other day here in chat.

The heavier the aircraft, the harder it is to stop.


So why isn't V1 reduced to allow for better stopping?
How is the max weight / V1 determined for a specific runway?




V1 is the speed by which the decision is made to continue or reject the takeoff.
Other than being controlled by Take-Off Run/Distance Available (TORA/TODA) or Accelerate Stop Distance Available (ASDA), all field limits, maximum $V_1$ is also controlled by $V_R$ (Speed for rotation) and $V_{MBE}$ (Maximum brake energy speed, but let's ignore $V_{MBE}$ in this context). $V_1$ must not exceed either.

$V_R$ increases with increasing gross weight.

This means that at low gross weight, theoretically, if the aircraft had not yet been rotated past a $V_1$ limited by $V_R$, sufficient distance may still exist to stop the aircraft on the remaining ASDA.

At higher gross weights, the higher $V_R$ required might allow for a higher $V_1$. Therefore, in response to: "So why isn't V1 reduced to allow for better stopping?": The higher $V_1$ still provides adequate margin for stopping on the ASDA, but gives you the option to abort the take-off at a higher speed with the increased $V_1$.

Rejected take-off (RTO) is still initiated when the engine failure is recognized, so a failure before the "increased" $V_1$ will result in RTO at the time of recognition of the failure, and not at $V_1$.

If performance allows, I prefer $V_1$ = $V_R$, as opposed to say $V_1$ = $V_R$ - 20 kt, on a short, slippery runway, which would imply that given an engine failure past $V_1$ you'd have to keep accelerating the aircraft with asymmetric thrust, on said runway, close to $V_{MCG}$ (minimum control speed on ground, which can be the lower limit for $V_1$).