Takeoff minimums, are you sure?

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Dave S.

Pre-takeoff checklist
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thetexan
we know that standard TO mims are 1 or 1/2 vis with expected climb gradient of 200 fpnm and a diverse turn no lower than 400 agl.

Let’s take another example. Fort Worth Borland.
STD/w min climb of 512’ per NM to 1300.

Question: what happens at 1300? Careful how you answer.

Look at the associated DP...Climb heading 165 to 1400 before turning right.

Consider this. A TO minimum seems logically to be a minimum set of conditions required to release the brakes for TO. The DP seems logically to be a set of instructions on how to fly the newly airborne aircraft after having released the brakes.

I talked to a POI at DAL FSDO as well as the individual at the operations group at ASW HQ iin Ft Worth that was responsible for the design of the area airport IAPs and neither gave the same answer and neither wa satisfactory.

If 400 ft in the standard TO min means you may now turn on course then 1300 should mean the same in the Bourland example. Then why 1400 in the DP? If you answer “...before turning right” then 1300 doesn’t really mean turn on just any old course as with the 400 example. It means turn on any course but right in which case you must go higher to 1400 before turning right.

If the TO min is really a minimum rather than a set of instructions then you would translate it like this....before releasing the brakes you must have either 1 or 1/2 mike viz, and your aircraft must be capable of climbing at 200 fpnm at least to 400. If that’s the case then you have your minimums and you can go. If not you have to go to an alternate minimum.

Anyway, The logic doesn’t follow from one minimum to another when DPs are involved.

So, what happens at 1300? Can I now stop climbing at 512 and go to 200 fpnm? Can I turn in any direction? What happens at 1400? Why the seemingly contradiction? If 1400 is the turn altitude then what is 1300 if not simply the altitude to which the cg must be maintained?

There is nothing in TERPs or 8260.46 that answers this.

Your thoughts
Tex
 
Were either of the answers given that you have to maintain 512 ft/nm to1300, at which time you can reduce to 200 ft/nm, but don’t make any right turns below 1400?

I’m not seeing a discrepancy there.
 
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Agreed. I don't see the contradiction. The 512 to 1300 is based on some obstacle that affects all routes. The 1400 is based on an obstacle that affects turning right. So you climb at 512 ft per nm to 1300, after which you climb at 200 ft per nm to whatever altitude you're cleared to. Once you get to 1400 MSL, if your route of flight is to the right, you can turn. If your route of flight is to the left, you can turn at 400 AGL which is the default. Presumably the obstacle driving the 1400 is the 1265 or 1372 MSL towers to the west of the airport.

You didn't copy the first part of the takeoff minimums, which is 300-2, which means if you have that ceiling and visibility, you can climb at 200 ft per nm all the way up. This allows you to see the obstacle causing the 512 ft per nm CG to 1300. However, if you are turning right, you still need to climb to 1400 before turning, due to (presumably) the towers to the west.

From the 8260.46, there are essentially three ways a procedure can be designed to avoid most obstacles:
1. Climb steeply enough to get over it.
2. If it's off to the side, climb high enough before turning, so you're over it.
3. Design a route that avoids the obstacle entirely.
Or a combination of the above. The .46 spells out all these options in Table 2-1-1, but understand its audience is not pilots, so there is a lot of background information and hunting around that is required to properly understand it.
 
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Don't use this procedure unless you can meet or exceed the required gradient to 1,300 feet. After that usual gradient 200 ft/nm) applies. Maintain 165 until 1,400. There is something in the way that makes this the safest course of action.

Remember this is based on ground speed, at 90 kt ROC has to be about 756 FPM, at 120 1,024. Loss of power in a single or an engine on a twin needs to be considered and may result in delaying departure or not going...if you are smart.

Hope this helps!

PS-Dave since you are a controller is there an FAA team that can answer questions about IFR routing? I'm planning a trip and want to take a short cut from a fix on a Victor airway to a fix on a different Victor airway to avoid climbing to a much higher MEA. Thanks!
 
I guess I should update my profile. I’m now retired from the FAA. I got out while I still have my sanity and am now training pilots in a college professional pilot program.

Tex
 
RussR said:
You didn't copy the first part of the takeoff minimums, which is 300-2, which means if you have that ceiling and visibility, you can climb at 200 ft per nm all the way up. This allows you to see the obstacle causing the 512 ft per nm CG to 1300.
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Keep in mind that the 200 ft/nm here is based on starting at 300 feet over the DER.
 
One of my issues is that some of these things are assumed and and written authority for it can’t be found. Or I can’t find it.

For example.

After The cg of 512 to 1300 we all assume we can return to 200 fpnm until reaching the enroute ROC. Where is that explicitly written.

It seems obvious that we don’t continue at 512 all the way up to enroute ROC; that at some altitude we clear the controlling obstacle and go back to using 200. But how
do we know we can return to 200? How do I prove to a persisting inquiring student that the non-standard cg ends at what we all assume is a turn altitude at 1300? How do we know that the 1300 doesn’t just apply to the turn. In fact how do we know it’s a turn altitude? We transfer that assumption from the fact that the example In the IPH states 400 as a turn altitude. But the IPH Says nothing about cg after that altitude mainly because the explanation in the IPH is using the standard 200/400 minimum. The assumption that cg returns to 200 after the altitude seems a good deduction but I can’t find anything written to prove it.

It also seems to me that if 1300 is not really a diverse turn altitude ( as evidenced by the 1400 right turn altitude) then 1400 should be used as the diverse altitude.

When I give talks in front of pilots they can ask very probing questions. Maybe I’m straining at a gnat.

Tex
 
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MauleSkinner said:
Keep in mind that the 200 ft/nm here is based on starting at 300 feet over the DER.
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No it's not. Where do you get that from?

The 200 ft per NM is calculated from the DER at the runway elevation for obstacle clearance purposes. The aircraft is assumed to cross the DER at 35 feet which allows for clearance over the takeoff obstacles in the "takeoff obstacle notes" but the actual surface is calculated from 0 AGL.
 
Dave S. said:
One of my issues is that some of these things are assumed and an written authority for it can’t be found. Or I can’t fine it.

For example.

After The cg of 512 to 1300 we can return to 200 fpnm until reaching the enroute ROC. Where is that explicitly written. How do we know we can return to 200? How do I prove to a persisting inquiring student that the non-standard cg ends at what we all assume is a turn altitude at 1300? How do we know that the 1300 doesn’t just apply to the turn. It seems a good deduction but I can’t find anything written to prove it.

It also seems to me that if 1300 is not really a diverse turn altitude ( as evidenced by the 1400 right turn altitude) then 1400 should be used as the diverse altitude.

When I give talks in front of pilots they can ask very probing questions.

Tex
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AIM 5-2-14 covers most of that, except that “we all” don’t assume that the 1300’ published is a turn altitude. Diverse departures would start at 1273 for anything but right turns. 1300 is just the top of where a steeper climb gradient is specified, and since a steeper climb gradient isn’t specified above 1300’, 200 ft/nm is required to the minimum IFR altitude
 
RussR said:
No it's not. Where do you get that from?

The 200 ft per NM is calculated from the DER at the runway elevation for obstacle clearance purposes. The aircraft is assumed to cross the DER at 35 feet which allows for clearance over the takeoff obstacles in the "takeoff obstacle notes" but the actual surface is calculated from 0 AGL.
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Obstacles don’t change because weather improves. As you indicated, the 300&2 allows some form of see & avoid, but the TERPS still have to address the obstacle, so the “300” is the altitude over the DER from which a 200 ft/nm gradient will clear the critical obstacle that requires the higher gradient from 35 feet.
 
Don't yet have the IR, just an IR student. I think it imprudent to take off into anything you can't get back into. That means for me the take-off minimums are going to look a whole lot like the landing minimums where I'm taking off, though if I got ILS I think I want it higher than that. I'd like to have the gear stowed before I hit the soup.
 
Dave S. said:
How do I prove to a persisting inquiring student that the non-standard cg ends at what we all assume is a turn altitude at 1300? How do we know that the 1300 doesn’t just apply to the turn. In fact how do we know it’s a turn altitude?
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Sometimes it really is just plain English. But you have to know the "defaults" first - meaning unless otherwise specified, you do "this". So, you must climb at 512 to 1300. After that there is no specific climb gradient requirement, so you revert to 200. If you are turning right, you must wait until 1400 to turn. If you are not turning right, you revert to the default which is 400 AGL. All of these are contained in various places, one of which is the AIM para 5-2-14.

You're not going to find an example spelled out the way you're looking for unless you go to a training presentation, which of course is what you are providing.
 
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steingar said:
Don't yet have the IR, just an IR student. I think it imprudent to take off into anything you can't get back into. That means for me the take-off minimums are going to look a whole lot like the landing minimums where I'm taking off, though if I got ILS I think I want it higher than that. I'd like to have the gear stowed before I hit the soup.
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That’s your prerogative, but you still need to determine how not to hit stuff.
 
MauleSkinner said:
Obstacles don’t change because weather improves. As you indicated, the 300&2 allows some form of see & avoid, but the TERPS still have to address the obstacle, so the “300” is the altitude over the DER from which a 200 ft/nm gradient will clear the critical obstacle that requires the higher gradient from 35 feet.
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While yes, being 300 feet above the DER and climbing at 200 ft per nm will certainly clear the obstacle, that's not what it means and not how it's calculated.

When a departure evaluation is run, a 40:1 surface is evaluated from the DER. If there are obstacles penetrating that surface, then an increased climb gradient is published. If those obstacles are located within 3 sm of the runway then a ceiling and visibility equal to the height above DER of that obstacle and distance to that obstacle (rounded up) is established. No further evaluation of anything like a 200 ft per nm climb gradient starting at that ceiling is accomplished. 8260.46G, Table 2-1-1 and others has this information.

Now, for a VCOA, your statement is largely correct, but that's not the scenario here.
 
MauleSkinner said:
Obstacles don’t change because weather improves. As you indicated, the 300&2 allows some form of see & avoid, but the TERPS still have to address the obstacle, so the “300” is the altitude over the DER from which a 200 ft/nm gradient will clear the critical obstacle that requires the higher gradient from 35 feet.
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No. To clear the low close in obstacle would require such an extreme cg (any cg over 500fpnm must be approved by fsdo) that a high cg solution is not practical. In that case, with low close in obstacles, a higher ceiling and increased viz solution is called for (8260.46). The idea being if the cg can’t practically be raised to clear the obstacle then you need a specified higher ceiling and viz so that the pilot can see and avoid the obstacle.

The obstacle clearance surface whether 40:1 or higher always begins at the DER height of 0 agl (sloping up from there) and is 76% of the total cg, the standard ROC being the other 24%. The standard takeoff minimum assumes the aircraft is at least 35 at the DER adding an additional 35 feet of buffer to the total clearance.

Tex
 
RussR said:
While yes, being 300 feet above the DER and climbing at 200 ft per nm will certainly clear the obstacle, that's not what it means and not how it's calculated.

When a departure evaluation is run, a 40:1 surface is evaluated from the DER. If there are obstacles penetrating that surface, then an increased climb gradient is published. If those obstacles are located within 3 sm of the runway then a ceiling and visibility equal to the height above DER of that obstacle and distance to that obstacle (rounded up) is established. No further evaluation of anything like a 200 ft per nm climb gradient starting at that ceiling is accomplished. 8260.46G, Table 2-1-1 and others has this information.

Now, for a VCOA, your statement is largely correct, but that's not the scenario here.
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Ok...guess I got that mixed up.
 
Dave S. said:
No. To clear the low close in obstacle would require such an extreme cg (any cg over 500fpnm must be approved by fsdo) that a high cg solution is not practical. In that case, with low close in obstacles, a higher ceiling and increased viz solution is called for (8260.46). The idea being if the cg can’t practically be raised to clear the obstacle then you need a specified higher ceiling and viz so that the pilot can see and avoid the obstacle.
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(deleted because I used the incorrect terms. Edited and reposted below.)
 
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What is the ICAO ID for this airport? It really helps everyone to post the ID when referring to a specific procedure or airport.
 
RussR said:
To be correct, the term "low close-in" applies to those obstacles within 3 sm which penetrate the 40:1 by less than 35 feet. These are the obstacles in the "takeoff obstacles" section of the departure procedure. A climb gradient is not calculated for these, nor is a ceiling and vis established to avoid them. They are there solely for pilot awareness, since a pilot crossing DER at or above 35 feet will clear them.
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That conflicts with the Instrument Procedures Handbook information, that says low, close-in obstacles are within 1 mile and less than 200 ft.
 
MauleSkinner said:
That conflicts with the Instrument Procedures Handbook information, that says low, close-in obstacles are within 1 mile and less than 200 ft.
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I misspoke (mis-typed?). And man I screwed up that post. "Low close-in" is a bit of an informal term, and I combined factors that confused the issue. There's a bit too much here for a brief discussion.

A "low close-in" obstacle, as defined by 8260.46G, is any obstacle within the "initial climb area" (basically straight out from the runway plus some to each side) that would cause a >200 ft per nm climb gradient but to only 200 feet or less above the DER. These obstacles are published as "takeoff obstacles" and the climb gradient to clear them is not published.

The "within 1 nm" is not explicitly specified in the 46G, however if it was more, the CG would be to an altitude higher than 200 ft above DER, so it's a correct statement. The statement "these obstacle are less than 200 feet above DER" is a simplication - true, they are all less than 200 feet above DER, but just because an obstacle is 199 feet above DER doesn't mean it's a low close-in - depends on more calculation than that.

Takeoff obstacle notes also include any obstacles within 3 sm that penetrate the surface.
 
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Dave S. said:
How do I prove to a persisting inquiring student that the non-standard cg ends at what we all assume is a turn altitude at 1300? How do we know that the 1300 doesn’t just apply to the turn. In fact how do we know it’s a turn altitude?
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1300 is NOT the turn altitude.
 
RussR said:
I misspoke (mis-typed?). And man I screwed up that post. "Low close-in" is a bit of an informal term, and I combined factors that confused the issue. There's a bit too much here for a brief discussion.

A "low close-in" obstacle, as defined by 8260.46G, is any obstacle within the "initial climb area" (basically straight out from the runway plus some to each side) that would cause a >200 ft per nm climb gradient but to only 200 feet or less above the DER. These obstacles are published as "takeoff obstacles" and the climb gradient to clear them is not published.

The "within 1 nm" is not explicitly specified in the 46G, however if it was more, the CG would be to an altitude higher than 200 ft above DER, so it's a correct statement. The statement "these obstacle are less than 200 feet above DER" is a simplication - true, they are all less than 200 feet above DER, but just because an obstacle is 199 feet above DER doesn't mean it's a low close-in - depends on more calculation than that.

Takeoff obstacle notes also include any obstacles within 3 sm that penetrate the surface.
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So the bottom line is 5hat simply crossing the DER at 35 feet and maintaining 200 ft/mile doesn’t ensure obstacle clearance on these?
 
steingar said:
You don't hit stuff by following departure instructions. I'm just referring to the weather.
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Your thinking about the weather minimums is correct. But an additional consideration is what are you going to do on take off roll-- in other words, what direction on take off, how high and how fast to climb, and when to turn and to what direction. The FAA IFR test is woefully inadequate on Obstacle Departure Procedures, in my opinion. They are not always assigned in your clearance. The AIM says that they will not assign it unless necessary for aircraft separation. But you should definitely know to check for them when taking off in IFR conditions, regardless of what your clearance is. According to the AIM, "If a Part 91 pilot is not given a clearance containing an ODP, SID, or radar vectors and an ODP exists, compliance with such a procedure is the pilot’s choice." But I cannot right now envision any reason not to fly the ODP if departing in IFR conditions.
 
MauleSkinner said:
So the bottom line is 5hat simply crossing the DER at 35 feet and maintaining 200 ft/mile doesn’t ensure obstacle clearance on these?
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Well no. It does when no obstacle penetrates the ocs. But it an obstacle does penetrate the ocs then, IF PRACTICAL, the ocs will be adjusted upward until the obstacle doesn’t penetrate. If the obstacle happens to be a low close in obstacle a ceiling/increased viz alternate minimum will be also be created.

The penetration of the ocs or lack of it drives everything.
 
Dave S. said:
Well no. It does when no obstacle penetrates the ocs. But it an obstacle does penetrate the ocs then, IF PRACTICAL, the ocs will be adjusted upward until the obstacle doesn’t penetrate. If the obstacle happens to be a low close in obstacle a ceiling/increased viz alternate minimum will be also be created.

The penetration of the ocs or lack of it drives everything.
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That’s what I thought. Just clarifying which parts of your earlier post to load shed. ;)
 
PPC1052 said:
1300 is NOT the turn altitude.
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What is that 1300 there for? What information does it convey in the takeoff minimum?

By the way, we’re using Bourland Field (50F) in our particular example.
 
Just for extra info, note the 4 nearest obstacles depicted on the RNAV 35 approach plate, closest to the runway. Those to the south, SE, and SW surely come into play for the obstacle clearance surface determination. Without looking up the exact formulae, I can guess the 1234' and 1372' might be where the 1300/1400 numbers come from.
 
Dave S. said:
What is that 1300 there for? What information does it convey in the takeoff minimum?
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That is the altitude to which you must* maintain the increased rate of climb in the event that the weather minimums are less than 300-2 but the weather meets the standard minimum for your aircraft. Once you have achieved that altitude, you may decrease your rate of climb to 200 ft/nm. If you can't maintain that rate of climb to that altitude, you can't depart IFR. Aircraft climb performance generally degrades with altitude, so it isn't out of the question that a particular aircraft can make that necessary climb gradient down low but can't at altitude. It makes sense that the FAA let's the pilot know the altitude at which they can safely reduce their climb so they can calculate based on conditions that whether they can make the necessary climb to that required altitude. But that 1300 number has nothing to do with making your turn.

The altitude that you may turn depends on your direction of turn. If to the right, its 1400. Otherwise, its the standard of 400 feet. You could get the clearance to depart 17, and turn left to heading 150. You would climb to 400, turn to 150, but you would want to keep your rate of climb up until you pass 1300, even though you made that turn at 400, so that you clear that power line and the towers to the south and to the east depicted on the VFR chart.




(*Of course, part 91 is not required to follow the alternative take-off minimums. So, from a legal requirement, we are only talking 135/121, etc. That being said, obstacles don't care if you are part 91, so it probably makes sense to observe these minimums, required or not. )
 
BOURLAND FIELD (50F)
TAKEOFF MINIMUMS AND (OBSTACLE) DEPARTURE PROCEDURES
AMDT 2 14261 (FAA)
TAKEOFF MINIMUMS: Rwy 17, 300-2 or std. w/min. climb of 512’ per NM to 1300.
DEPARTURE PROCEDURE: Rwy 17, climb heading 165° to 1400 before turning right.
TAKEOFF OBSTACLE NOTES: Rwy 17, trees beginning 9’ from DER, 87’ right of centerline, up to 60’ AGL/886’ MSL.....

Ok, I am flying an aircraft incapable of 512 per NM, using 300-2 doesn’t do anything for obstacle clearance. The obstacle procedure is climb to 1400 before turning RT.

The runway 35 elevation is 865. 1300 is a 435 ft above airport And requires a .8 NM climb.
 
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Clip4 said:
Ok, I am flying an aircraft incapable of 512 per NM, using 300-2 doesn’t do anything for obstacle clearance.
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True. But it does let you see the obstacle so you can avoid until you are high enough to be clear.
 
Dave S. said:
If the obstacle happens to be a low close in obstacle a ceiling/increased viz alternate minimum will be also be created.
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Low close-in obstacles don't trigger a ceiling publication unless the DER is below field elevation. Here's a tutorial of mine on the subject at hand:
http://www.avclicks.com/Flash2/Stairway_to_Heaven2/index.html
http://www.avclicks.com/Flash2/Stairway_to_Heaven2/index.html

It is "implied" that standard minimum 200 feet ceilings are assumed for all IFR departures.
 
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Clip4 said:
The runway 35 elevation is 865. 1300 is a 435 ft above airport And requires an .8 NM climb.
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Maybe Russ can explain this, because I am not certain. I suppose that they let you fly with standard climb gradient under the 300-2 minimuns, which potentially puts you closer to obstacles than the 1300 with the increased climb gradient, because you are see and avoid rather than relying on TERPS standards for clearance.
 
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