Calculating Climb Gradient

Jaybird180

Jaybird180

Final Approach
Joined
Dec 28, 2010
Messages
9,036
Location
Near DC
Display Name
Display name:
Jaybird180
Suppose a DP requires a non-standard climb. If I know my climb performance at 90kts gets me 600FPM, how do I calculate the climb gradient for obstacle clearance?

My HS trig tells me that I need to know the angle of the climb. Or am I overthinking?
 
Jaybird180 said:
Suppose a DP requires a non-standard climb. If I know my climb performance at 90kts gets me 600FPM, how do I calculate the climb gradient for obstacle clearance?

My HS trig tells me that I need to know the angle of the climb. Or am I overthinking?
Click to expand...
No trig, just arithmetic. Climb gradient is feet/nm. Climb rate is ft/min. Ground speed is nm/hr. GS/60=nm/min. Climb rate divided by GS/60 gives gradient in ft/nm.

(ft/min)/(nm/min) = ft/nm.

Just remember to use GS rather than IAS in your calculations -- IAS converted to TAS +/- wind and all that.
 
Where do you find the GS for any given IAS. For expample, reviewing the POH performance section for the DA-40 I was unable to ascertain this information.
 

Attachments

  • Climb-Chart.jpg
    Climb-Chart.jpg
    57.7 KB · Views: 32
Jaybird180 said:
Suppose a DP requires a non-standard climb. If I know my climb performance at 90kts gets me 600FPM, how do I calculate the climb gradient for obstacle clearance?

My HS trig tells me that I need to know the angle of the climb. Or am I overthinking?
Click to expand...

The angle can give you what you desire, or you can know the distance to the obstacle and its height and then figure if you will make that height by the time you make that distance.
 
skidoo said:
The angle can give you what you desire, or you can know the distance to the obstacle and its height and then figure if you will make that height by the time you make that distance.
Click to expand...
The DP may not tell you exactly where the limiting obstruction is or its exact height, or the climb angle, only the climb gradient (in ft/nm) you need to fly the DP without hitting anything. See AIM 5-2-8 for details.

If you want to know the climb angle (not sure what utility it would have), you could take the gradient in ft/nm, divide that by 6076 ft/nm, and the resulting number will be the tangent of the climb angle -- you do remember Chief SOH-CAH-TOA of the Geometry tribe of Indians (sorry -- "Native Americans"), right?

[we can talk about his three squaws and right triangles another time, but if you can't wait, click here]
 
Last edited:
Let me ask it this way...(then you can correct me later).

Provided in the OP is the velocity the ship is traveling up the hypotenuse of the triangle. The Rate provides the height (vertical standing leg) of this right triangle, if one knows how far up the hypotenuse the ship travels in one minute. From this, the angle can be ascertained.

Provided that this angle exceeds the required angle, the performance requirements of the DP are said to have been met. If not, do not use that DP.
 
Last edited:
Ron Levy said:
The DP may not tell you exactly where the limiting obstruction is or its exact height, or the climb angle, only the climb gradient (in ft/nm) you need to fly the DP without hitting anything. See AIM 5-2-8 for details.

If you want to know the climb angle (not sure what utility it would have), you could take the gradient in ft/nm, divide that by 6076 ft/nm, and the resulting number will be the tangent of the climb angle -- you do remember Chief SOH-CAH-TOA of the Geometry tribe of Indians (sorry -- "Native Americans"), right?

[we can talk about his three squaws and right triangles another time, but if you can't wait, click here]
Click to expand...

Where did this number come from? I think we need to have that triangle discussion.
 
Jaybird180 said:
Let me ask it this way...(then you can correct me later).

Provided in the OP is the velocity the ship is traveling up the hypotenuse of the triangle. The Rate provides the height (vertical standing leg) of this right triangle, if one knows how far up the hypotenuse the ship travels in one minute. From this, the angle can be ascertained.

Provided that this angle exceeds the required angle, the performance requirements of the DP are said to have been met. If not, do not use that DP.
Click to expand...
You can do it that way if you want, but I think my way (converting required gradient directly to minimum climb rate without bothering with converting to angles) is a lot easier. I suspect more folks remember their 4-function arithmetic than those who remember their trig, but YMMV.
 
Using either method I am unable to ascertain if the airplane I am flying meets the climb gradient requirements.

Can you show me an example using the performance page posted?
 
OK, let's assume you're at max gross from Jackson Hole, WY (KJAC) on the Geyser Four at MGW (1150 kg/2535 lb) and climbing at Vy (73 KIAS). Book climb rate from the chart you posted at the field elevation of 6451 (assume standard day, temp +2C) is 520 ft/min. TAS at 73 KIAS at 6451 MSL at +2C spun off the whiz wheel is 80 knots. Now we spin 80 knots TAS and whatever the wind is on the other side of the whiz wheel; say it comes out 75 knots. Actual initial climb gradient is then 520/(75/60) [see formula posted above] is 416 ft/nm, and goes down as you climb. Required climb gradient off the DP chart is 450 ft/nm all the way to 14,000. 416<450 ==> no can do.
 
Jaybird180 said:
Where did this number come from? I think we need to have that triangle discussion.
Click to expand...

It came from whoever decided how long a nautical mile is.

Jay you are over thinking this.

Ron's approach approximates the climb gradient, assuming that the hypoteneuse is appriximately equal to the base of the triangle. This is a good approximation when the angles are small. Which for GA airplanes the climb angle is small, generally <10 degrees.

You're approach will result in the exact answer, but requires a calculator.
 
COFlyBoy said:
It came from whoever decided how long a nautical mile is.

Jay you are over thinking this.

Ron's approach approximates the climb gradient, assuming that the hypoteneuse is appriximately equal to the base of the triangle. This is a good approximation when the angles are small. Which for GA airplanes the climb angle is small, generally <10 degrees.

You're approach will result in the exact answer, but requires a calculator.
Click to expand...

Okay, I can get with that. Why didn't you just say so:lol::thumbsup:
 
How much margin would be prudent to add to a DP?
 
Jaybird180 said:
How much margin would be prudent to add to a DP?
Click to expand...

I wouldn't add any, since almost any plane you'll be flying can meet the required climb gradient (unless you're starting at a high altitude) and the DP already has margin built into it.

Interestingly some of the high performance singles we all drool over, don't have as steep of a climb gradient as a Skyhawk. I recall reading that the Corvallis gradient is shallower than the Skyhawk, even though the climb rate is higher.
 
Ron Levy said:
You can do it that way if you want, but I think my way (converting required gradient directly to minimum climb rate without bothering with converting to angles) is a lot easier. I suspect more folks remember their 4-function arithmetic than those who remember their trig, but YMMV.
Click to expand...
I'm with you here, including using the E6B. :D
And how quickly one covers distance along the hypotenuse is irrelevant, as is the angle... in fact, except in perfectly still, consistently "standard" air with a good autopilot flying, neither of those will be constant anyway. The other stuff will probably not be constant either, but we need to work with something, and we're not planning to clear said obstacle without some extra room, just in case.
All that matters is time to reach X altitude, flying a specific heading, given all factors including wind, and the horizontal distance derived from the groundspeed, once the time is determined. I mean, that's where the obstacle is: a given horizontal distance away on a specific heading, and a given vertical distance from the start-of-climb altitude.

I don't even sweat the gradient per se, just the result: "Obstacle is X ft MSL (including a safety margin); at Vy (under these conditions) it will take XX minutes to reach that altitude... distance (horizontally) to obstacle is XX miles; at my estimated GS it will take XX minutes to cover that distance. Can I make it? No? OK, then try it at Vx, or maybe plot two or more climbs (with a turn somewhere), or plan to climb in a circle for XX minutes then head towards the obstacle on my course. Can I make it (at Vy) it in a straight line with more extra altitude than I need? Okay, then let's try a cruise climb, to avoid wasting time (and distance) climbing."

Other than the climb performance numbers and the crosswind component, it's pretty simple skull-work, even for a math-impaired pilot such as myself.

Then there's usually a "put slight dogleg in course to avoid all that stuff" option... :D
 
Last edited:
COFlyBoy said:
I wouldn't add any, since almost any plane you'll be flying can meet the required climb gradient (unless you're starting at a high altitude) and the DP already has margin built into it.

Interestingly some of the high performance singles we all drool over, don't have as steep of a climb gradient as a Skyhawk. I recall reading that the Corvallis gradient is shallower than the Skyhawk, even though the climb rate is higher.
Click to expand...

fully loaded C-421 had about the same climb angle as a C-150 in my experience. The numbers were just doubled, say 120 mph IAS and 5-600 fpm. only difference was it could carry a few thousand more pounds...
 
Jaybird180 said:
Suppose a DP requires a non-standard climb. If I know my climb performance at 90kts gets me 600FPM, how do I calculate the climb gradient for obstacle clearance?

My HS trig tells me that I need to know the angle of the climb. Or am I overthinking?
Click to expand...

No, you do not need trig unless you want to calculate the angle in degrees (or radians). (Can be used for calculating descent rates for glideslopes, which are given in degrees, though).

You're trying to convert it the opposite way it's usually done. You certainly can do it that way, but usually you take the climb gradient in feet per mile and convert it to the required fpm at your groundspeed. I think it's done that way because it's easier to divide by 60 and multiply by some number in your head than it is to divide by 60 and then divide by some number.

Remember a knot is just a nautical mile/hour.

gs in nm/hr
----------- * gradient in ft/nm == climb rate in ft/min
60min/hr

The hours cancel out, the nautical-miles cancel out, and you get your minimum feet per minute climb rate.

Remember multiplication and division get the same priority in the order of operations so it doesn't matter which order you do the calculations in.

Your example, working the opposite way:

90 nm/hr
--------- * gradient = 600ft/min
60min/hr

1.5nm/min * gradient = 600ft/min

gradient = (600 ft/min) / (1.5nm/min)
gradient = 400ft/nm
 
Last edited:
Simple solution: use the climb/descent table in the back of the FAA TERPS plates, or in the Jepps.
 
You must log in or register to reply here.
Back
Top