Takeoff minimums, are you sure?

aterpster said:
Except we are not required to turn higher than 1,200, but we must make good the applicable climb gradient to either 3,400 or 4,600.
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That's not an except. We've all been saying that. Except Dave who believes the climb gradient to altitude requirement is a turn restriction. I used the examples only to show that if Dave was right, there was a conflict between the climb gradient and the ODP. Of course, there is no conflict because they deal with different things.
 
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DogoPilot said:
What I gather from the responses and from reading the procedure in the first post is that you're safe to start a turn to the left at 400' but you have to continue to climb at the higher climb gradient up to 1300' (not sure where the 1100' is coming from). Turning and climbing at the higher gradient aren't mutually exclusive.

I think the example given by denverpilot makes it clear why this could be the case.
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DP50F.JPG
You "gather" correctly, except you aren't necessarily "safe" until you are above 1300'. Below that altitude you must visually avoid obstacles even after turning at 400' AGL. The significance of 1300' is the diverse sector drawn from the runway has obstructions penetrating the 40:1 plane below that altitude. You need higher than a 200' AGL ceiling to see and avoid the top of them or you need 512'/nm to top them by 35' with standard (1 mile) visibility. If there are no departure route restrictions (here, there is for right turns only) you can turn on course as early as 400' AGL and visually dodge them up to 1300'.
 
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MauleSkinner said:
Because there is more than one obstacle in the vicinity of the airport for which the gradient is required. You’re not “safe” as the term is used in the question.
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We’re making progress? So please explain the takeoff minimum process. Remember no DP.

Based on F53 238 to 900. Which way can I turn at 400 and which direction is it I have to wait to 900 to turn to based on the different obstacles?

I believe what you are doing is making up your own DPs based on where the obstacles are. Which is fine. You’re free to do that. But that doesn’t deal with why the FAA has 238 to 900.

238 to 900 is the FAA saying that if you want to be able to turn in any direction without worrying about where the obstacles are then we (the FAA) can only guarantee that at 900. And only if you climb at 238 fpnm. They also allow (by not making a prohibition against it) for you to turn any which way you want by using the obstacle data to do so.
 
midlifeflyer said:
A FSDO answer is not the FAA's answer.
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FSDO POI has the authority. Like a MLB umpire. The POi for your operation is your authority. Who are you kidding. Now you’re showing your ignorance. I’ve worked closely with FSDO for 30 of my 37 year in the business
 
Dave S. said:
We’re making progress? So please explain the takeoff minimum process. Remember no DP.

Based on F53 238 to 900. Which way can I turn at 400 and which direction is it I have to wait to 900 to turn to based on the different obstacles?

I believe what you are doing is making up your own DPs based on where the obstacles are. Which is fine. You’re free to do that. But that doesn’t deal with why the FAA has 238 to 900.

238 to 900 is the FAA saying that if you want to be able to turn in any direction without worrying about where the obstacles are then we (the FAA) can only guarantee that at 900. And only if you climb at 238 fpnm. They also allow (by not making a prohibition against it) for you to turn any which way you want by using the obstacle data to do so.
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If you’re going to maintain your incorrect assumptions, my answers aren’t going to make any more sense to you than they did when I answered them before.

It’s like you’re asking, “since the world is flat, how far away is the edge?” and expecting an actual distance.
 
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Dave S. said:
FSDO POI has the authority. Like a MLB umpire. The POi for your operation is your authority. Who are you kidding. Now you’re showing your ignorance. I’ve worked closely with FSDO for 30 of my 37 year in the business
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POI has authority over a specific operation. Doesn't mean they know the rules any more than a cop walking the beat knows the Constitution.
 
dtuuri said:
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You "gather" correctly, except you aren't necessarily "safe" until you are above 1300'. Below that altitude you must visually avoid obstacles even after turning at 400' AGL. The significance of 1300' is the diverse sector drawn from the runway has obstructions penetrating the 40:1 plane below that altitude. You need higher than a 200' AGL ceiling to see and avoid the top of them or you need 512'/nm to top them by 35' with standard (1 mile) visibility. If there are no departure route restrictions (here, there is for right turns only) you can turn on course as early as 400' AGL and visually dodge them up to 1300'.
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I guess I’m confused now. I assumed that in the scenario where you were operating under the “std. w/min climb of 512’ per NM to 1300” we wouldn’t be able to avoid obstacles visually because the ceiling could theoretically be 0 AGL. Is that incorrect?
 
DogoPilot said:
I guess I’m confused now. I assumed that in the scenario where you were operating under the “std. w/min climb of 512’ per NM to 1300” we wouldn’t be able to avoid obstacles visually because the ceiling could theoretically be 0 AGL. Is that incorrect?
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The last thing I want to do is confuse somebody. "Std" means one or one-half mile visibility, depending if single or multi. If you looked at my tutorial you can see how a 200' ceiling is implied as part of standard minimums — because of low close-in obstacles and the need to visually avoid them. They only publish higher than 200' ceiling minimums. Doesn't make sense, to me, to not have any ceiling minimum at all for the most critical of all (close-in) obstacles. Not saying it applies to air carrier ops, but for 91 operators formulating good operating procedures I think it should be a consideration.

Theoretically, the "512'/nm" should clear obstacles if you turn at 400' AGL and don't turn right i/a/w the route departure restriction, but you're climbing out in an uncharted forest of obstructions poking through the shallow 40:1 plane. And you're having to keep a much steeper gradient than that standard plane. Any mis-fire in performance could have tragic consequences. So, it isn't "safe" to assume you will really clear all those hidden protrusions out there in the murky sky without some semblance of visibility and ceiling enough to avoid them visually. So, I'm saying until you regain the clean floor of the diverse sector where your normal climb gradient far exceeds the 200'/nm assumed performance — you should plan on being able to avoid the obstacles visually and not rely on theory.
 
dtuuri said:
The last thing I want to do is confuse somebody. "Std" means one or one-half mile visibility, depending if single or multi. If you looked at my tutorial you can see how a 200' ceiling is implied as part of standard minimums — because of low close-in obstacles and the need to visually avoid them. They only publish higher than 200' ceiling minimums. Doesn't make sense, to me, to not have any ceiling minimum at all for the most critical of all (close-in) obstacles. Not saying it applies to air carrier ops, but for 91 operators formulating good operating procedures I think it should be a consideration.

Theoretically, the "512'/nm" should clear obstacles if you turn at 400' AGL and don't turn right i/a/w the route departure restriction, but you're climbing out in an uncharted forest of obstructions poking through the shallow 40:1 plane. And you're having to keep a much steeper gradient than that standard plane. Any mis-fire in performance could have tragic consequences. So, it isn't "safe" to assume you will really clear all those hidden protrusions out there in the murky sky without some semblance of visibility and ceiling enough to avoid them visually. So, I'm saying until you regain the clean floor of the diverse sector where your normal climb gradient far exceeds the 200'/nm assumed performance — you should plan on being able to avoid the obstacles visually and not rely on theory.
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Thanks for the explanation and link to the tutorial. As a newly rated instrument pilot, my personal takeoff minimums far exceed the standard takeoff minimums or any non-standard takeoff minimums so I am just trying to make sure I understand it from an academic perspective based on the context of the example in the OP.

It is my understanding that the minimum climb gradient requirements to 1300' and the altitude at which you can start a turn are completely independent of each other; however, it seems that the OP is assuming that some dependency exists and I believe this is incorrect. Maybe "safe" wasn't the best choice of words in my first post, but is it correct that if the weather is 200-1 (implying that visual separation cannot be maintained from obstacles more than a couple of miles out) then by climbing at a 512'/nm gradient to 1300', regardless of whether on runway heading or turning to the left at 400' AGL, obstacles in the vicinity of this airport would be cleared?

EDIT: It just registered in my mind that you already answered my question in your post, so no need to reiterate.
 
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DogoPilot said:
It is my understanding that the minimum climb gradient requirements to 1300' and the altitude at which you can start a turn are completely independent of each other; however, it seems that the OP is assuming that some dependency exists and I believe this is incorrect.
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You are correct.
 
DogoPilot said:
EDIT: It just registered in my mind that you already answered my question in your post, so no need to reiterate.
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Looking at the airport diagram for 50F, the departure end of RWY 17 is .9% below the field elevation, or 836 ft. That is 210 feet below the obstruction charted SE of the airport. The required ceiling and visibility for departure is 300 & 2, which makes sense because they round up. If the obstruction were 11 feet lower, since the ceiling would be 200, it would not be published, and, I imagine, dictate just a standard takeoff visibility of one mile (SEL). I'd want to have 200 vertical feet of one mile visibility before takeoff, at a minimum, even if my climb gradient on paper seems it should clear that tower.
 
MauleSkinner said:
Yup...I was just noticing the pattern. (Some days I’m a little slow. :oops: )
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Well, first let me apologize for what I said to you. I was harsh.

Second I have spent all day researching this as I try to prepare a lesson plan on this subject. I found ac 120-91 which made it finally click about what you were trying to explain. I was hearing one thing and you were saying another.

When you take into considerations of weight restricted takeoffs, minimums and the like I think I understand what I thought was a conflict of logic.

Again thanks
 
I have PTSD from doing TERPS approach/containment diagrams.

No human should be subjected to such inhumane torture....
 
Dave S. said:
I have spent all day researching this as I try to prepare a lesson plan on this subject. I found ac 120-91 which made it finally click...
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Boy, there's a non sequitur if ever there was one. AC 120-91 explains one engine inoperative procedures in lieu of TERPS which are based on normal ops and is what your student is asking about.
 
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dtuuri said:
Looking at the airport diagram for 50F, the departure end of RWY 17 is .9% below the field elevation, or 836 ft. That is 210 feet below the obstruction charted SE of the airport. The required ceiling and visibility for departure is 300 & 2, which makes sense because they round up. If the obstruction were 11 feet lower, since the ceiling would be 200, it would not be published, and, I imagine, dictate just a standard takeoff visibility of one mile (SEL). I'd want to have 200 vertical feet of one mile visibility before takeoff, at a minimum, even if my climb gradient on paper seems it should clear that tower.
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https://nfdc.faa.gov/nfdcApps/services/ajv5/airportDisplay.jsp?airportId=50F
 
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