As a former performance engineer for Cessna, and having been around during CE-560 certification, I can definitively tell you, it does have to do with certification requirements. (V1min, and hence V1 not being lower than Vr)

As a former performance engineer for Cessna, and having been around during CE-560 certification, I can definitively tell you, it does have to do with certification requirements. (V1min, and hence V1 not being lower than Vr)

I get the V speed minimums, but still, if you are 1000 lbs lighter, and use the SAME V speeds, then airplane will use less runway. The book doesn't agree with that so the numbers are no longer actually tied to aircraft performance at that point.

The only way the book numbers reflect true performance is if the engine somehow develop less power when the airplane is lighter, which is nonsense.

Mike C.

MCT = maximum cruise thrust. I used the wrong words above, edited.

LRC = long range cruise.

MCT and LRC define the boundaries is useful power settings in cruise. MCT is as fast as you can go. LRC is a slow as you can go without lowering range.

Mike C.

Mike - please show your math on that 1300 mile leg into 100kts of wind. I am showing the ultra burns a lot more fuel. Am I missing something? V can’t be that much more efficient.

I was using FF profiles, max cruise on ultra, cxy bjc, fuel stop for mu2 in Omaha.

It's an oldy but a goody...

https://www.youtube.com/watch?v=px8awNCvQFM

What’s Vmca and Vmcg, this limit can determine V2

Andrew

I get the V speed minimums, but still, if you are 1000 lbs lighter, and use the SAME V speeds, then airplane will use less runway. The book doesn't agree with that so the numbers are no longer actually tied to aircraft performance at that point.

The only way the book numbers reflect true performance is if the engine somehow develop less power when the airplane is lighter, which is nonsense.

Mike C.

I found this article with an interesting discussion:

https://aviation.stackexchange.com/ques ... g-distance

It's noted that in theory, braking distance from the same velocity is independent of mass as long as the brakes are not energy-limited.

However, a heavier aircraft landing at the same speed (or for our purposes, we could say aborting at the same speed) as a lighter aircraft may actually be able to stop in a shorter distance when residual lift is factored in because "the higher landing weight keeps a higher downforce percentage after deploying spoilers..."

They are the same whether the plane weighs 11,000 lbs or 12,000 lbs.

The 11,000 lbs plane reaches V1 quicker than the 12,000 lbs plane. Even if it takes the same distance to stop, the 11,000 plane will use less runway.

Given the same airspeeds for V1, Vr, the 11,000 lbs plane is suffering the same air drag and speedbrake drag. Thus that will slow it down faster than the 12,000 lbs airplane even if the brakes are not as effective with a lighter weight.

I can tell you from direct observation that the 11,000 lbs plane will stop quicker. Indeed this is reflected in the landing distance numbers:

ISA, 0 MSL, no wind:

11,000 lbs: 2150 ft
12,000 lbs: 2310 ft

So the theory that the lighter weight doesn't reduce braking distance is bunk.

Also, why does takeoff distance reduce between 15,900 lbs and 13,000 lbs, but then stops reducing below 13,000 lbs? Clearly weight reduction in that region makes a difference.

The numbers have been tampered with, plain and simple. There's no regulatory or physics reason the 11,000 lbs plane has the same performance as the 12,000 lbs plane.

Mike C.

Mike, I tried to explain above.
The reason is likely acceleration through V1 during an AEO abort, during the action completion, and V1 not getting any lower with lower mass due to VMCA or VMCG.

Lighter plane will reach V1 in less runway even when V1 is the same as the heavier airplane. After that, the lighter airplane will not take longer to stop.

Overall, lighter plane stops in less runway even when the abort portion of the exercise takes the same distance.

The pre V1 segment of the takeoff is with the same thrust and all engines. It simply can't be the lighter airplane takes the same distance to reach V1.

Mike C.

You‘ll exceed V1. Abort is about delay times. Again: depending on certification standards. Your AFM might give a hint, but won‘t show all.

https://code7700.com/v-1.htm#gsc.tab=0


That‘s obvious, even for lift on ground roll: VREF prop. SQR(m)
(And the lift will be the same percentage of mass on touchdown for different masses.)

It doesn’t take a rocket scientist to figure that out.

My bet is that once they hit Vmca/mcg, Cessna stopped calculating and locked the speeds / TOD at that point. Who would want data below 2300 any way.

Andrew

It‘s not necessarily that easy, that they just don‘t calculated.
On the Mustang, e.g., when being very light and you do use Flaps 15 (=> lift on ground roll vs. less, no or negative lift with Flaps Up), you might even get a higher required TOD than at a higher mass (see image below, numbers in red circles). With Flaps Up, it‘s vice versa.

The multiple reasons for this can be read in my posts above.
Again, top speed during an AEO V1 cut will be well above V1, the lighter the more, and deceleration can be less at lower mass if there‘s lift.

It‘s not rocket since, but it‘s a = a (v, m) with a(v) not being available to us and delays during action completion time and spool down.
What‘s the angle of attack on ground at what flap setting?

I‘d trust the AFM.