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Requesting for some clarification. Part 91 operator. 0-0 visibility and ceiling minimums

Departing Rwy 15 at ADS

Can I depart with a standard 200 ft/nm cross DER 35 ft. since I don't need to comply with the 400-2 1/2 wx? (0-0 ...right?)

or

Do I fall under the std.(which from what I understand, is ...0-0?) w/ min. climb of 325' per NM to 1100?

ADS

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    $\begingroup$ I would never take off from someplace where I couldn't land in case there was some problem. 0-0 visibility would keep me on the ground until it cleared up some. $\endgroup$
    – CrossRoads
    Jul 30, 2018 at 18:03

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What this tells me is that I can take off with any legal visibility/ceiling for my type of operation if I can maintain 325’ per nm climb. If I can only accomplish the standard 200’ per nm climb, I need 400’ ceilings and 2.5 miles of visibility, regardless of operation type.

Essentially it’s saying that there’s something out there that you’ll hit if you can’t maintain the 325’ per nm climb gradient. If you can, great, go. If not, you need the increased visibility to see and avoid it.

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You are incorrect in your understanding of standard. According to

§91.175 Takeoff and landing under IFR. (f) (2) (i) For aircraft, other than helicopters, having two engines or less—1 statute mile visibility. (ii) For aircraft having more than two engines— 1⁄2 statute mile visibility.

However, these minimums do not apply to you because just before the definition of standard it says:

(f) Civil airport takeoff minimums. This paragraph applies to persons operating an aircraft under part 121, 125, 129, or 135 of this chapter.

So standard minimums are 1 statute mile for 1 and 2 engine aircraft and 1/2 statute mile for more than two engines but the rule doesn’t apply to you since you are flying under Part 91.

The next part of that section says:

(3) Except as provided in paragraph (f)(4) of this section, no pilot may takeoff under IFR from a civil airport having published obstacle departure procedures (ODPs) under part 97 of this chapter for the takeoff runway to be used, unless the pilot uses such ODPs or an alternative procedure or route assigned by air traffic control.

In this case, since there is a published ODP, you are required to have either the visibility or the climb performance. If you can climb at 325' per nm to 1100' then you can take off in 0/0 conditions.

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  • $\begingroup$ What about the 400-2 1/2 wx? is that a minimum climb of 200ft/nm? $\endgroup$
    – RAU
    Jul 31, 2018 at 5:02
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    $\begingroup$ @RAU The visibility minimums are there so you don’t hit something that you can’t see. Presumably you’ll be able to see the obstacle and maneuver around it. So a climb rate of 200 ft/nm would be fine. If there is rising terrain that requires a climb rate of more than standard, then you would have a notation that requires a specific climb rate. $\endgroup$
    – JScarry
    Jul 31, 2018 at 15:28
  • $\begingroup$ I thought F(3) does not apply to part 91? "(f) Civil airport takeoff minimums. This paragraph applies to persons operating an aircraft under part 121, 125, 129, or 135 of this chapter." $\endgroup$
    – RAU
    Jul 31, 2018 at 20:47
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    $\begingroup$ @RAU Exactly. The definition of standard takeoff minimums is the same for everyone. But only categories listed in (f) (3) need to comply with the minimums defined in (f) (2) (i). $\endgroup$
    – JScarry
    Jul 31, 2018 at 21:42
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    $\begingroup$ @RAU You have two choices listed on the departure procedure. You can depart with 400 and 2 1/2 and 200 ft/nm with 35' above the departure end of the runway OR you can depart with 0/0 visibility (since you are Part 91) and 325' per NM up to 1100' . $\endgroup$
    – JScarry
    Aug 1, 2018 at 1:14

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