Part135 Guy

If you've flown large transport category a/c under 135, we all know for destination airport runway length calculations we all take our AFM landing distance for that weight and temp and divide by 60 to meet the following regulation:

135.385(b) Except as provided in paragraph (c), (d), (e), or (f) of this section, no person operating a turbine engine powered large transport category airplane may take off that airplane unless its weight on arrival, allowing for normal consumption of fuel and oil in flight (in accordance with the landing distance in the Airplane Flight Manual for the elevation of the destination airport and the wind conditions expected there at the time of landing), would allow a full stop landing at the intended destination airport within 60 percent of the effective length of each runway described below from a point 50 feet above the intersection of the obstruction clearance plane and the runway.

Futher, 135.361(b) gives us guidance on the definition of effective runway length:

135.361(b) For the purpose of this subpart, effective length of the runway, for landing means the distance from the point at which the obstruction clearance plane associated with the approach end of the runway intersects the centerline of the runway to the far end of the runway.

Read more: http://cfr.vlex.com/vid/135-385-turbine-powered-landing-airports-19561847#ixzz0oDdKLsRj

Someone, please, with obviously more mathematical prowess than I posess, explain to me why we divide the AFM landing distance number by sixty when it looks to me, from what I read of black print on white paper, we should be finding 60 percent of the published runway length and then comparing that number to our AFM landing distance.

PW305

6000' runway... must be able to stop in 3600' by the AFM

or take 3600' from the book and divide by 0.60 = 6000'

not sure I understand the question

Part135 Guy

Lets say you have a 5000' runway, AFM lists runway required as 2500'.

2500/60=4166'

But the regs say to take 60% of 5000

5000 x .6 = 3000'

Picking up what I'm putting down?

PW305

2500 is less than 3000

4166 is less than 5000

you can legally use the runway

the 500' difference in unfactored distances turns into a difference of 833' when factored which is, surprisingly, 60% higher

Part135 Guy

Soooooooo.... which number represents required runway length for landing? Is the required length 3000' with 2000' to spare or 4166' with 833' to spare? It seems in your response its either/or.

Or better asked, which number will my FSDO appointed check airman want to see on my type ride?

PW305

You're confusing yourself! Required runway length and available landing distance are exclusive of each other. If the question is how much runway is required, the answer is 4166'... provided unfactored AFM number is 2500', you're operating as a part 135 (without at DAAP and 80% approval) and the runway is dry.

It's 4166' whether landing distance available is 5000' or 10000'

Part135 Guy

No, I understand the difference between runway required and available runway.

What I AM confused about is why do we divide 60% into our AFM landing distance number? This is increasing the number in the AFM. The regs don't say to do this. The regs clearly state that we are to land and stop within 60% of the effective runway (essentially runway length minus the distance covered where the obstruction clearance plane and runway centerline meet). So, why then, do we not take the effective runway length and multiply by .6, compare that number to the AFM performance number to see if we're legal?

If you divide .6 into your landing distance, what you are doing is saying, ok, for every foot of landing distance my AFM says I need, I'm going to add 2/3 or .666 feet to that foot. This is how we arrive at 4166 from an AFM value of 2500. But the regs don't say to do this. This is why I am confused.

GauleyPilot

Part 135 guy, it comes down to two different ways to figure it.

Choice 1. The one you mentioned that fits with the reg better.

Example: 5,000 ft rwy.

AFM: 3200 feet required.

5000 X.6 = 3,000 feet.

You are 200 feet over on AFM runway required, and can not do it.

Choice 2. The AFM/ .60 way.

5,000 ft rwy.

AFM: 3,200 feet required

3,200/.60 = 5,333 feet. You have 5,000ft of runway available. You cant do it. You need 333 more feet.

Think of it as one telling you what you will get, and one telling you what you will need.

Take a 5334 ft rwy now using Choice number 1.

5334 X .60 = 3200 feet.

AFM requres 3200 feet. You can do it.

GauleyPilot

In this example, you are ok because you only need 2,500 feet to stop. That is within the bounds of the 3000 feet required in 60% of a 5,000 ft. rwy.

You could do it on a 4166ft rwy if you wanted to also, because 2,500 is 60% of that.

Two methods, one result.

PW305

why do we divide 60% into our AFM landing distance number?

Because that is how you work the problem backwards, to make it easier. But you don't have to do it that way... follow me through

Forget about any specific airport or runway length. Lets talk temp & field elevation, and say your unfactored landing distance is 2500'. Since you must stop within 60% of some runway, what is the minimum length you can accept? 4167', we've answered that.

Lets work it the other way... you want to land at an airport with a 7000' runway. You can either go to the book and look at the factored distance and see if it's equal to or less than 7000'. OR you could find 60% of 7000', which is 4200' and see if the unfactored distance required is less.