General Principles - Climb — Page 195, Lesson 225

Let’s pick this up right where the climb gradient work left off. We’ve already got the air gradient — the climb angle the aeroplane achieves through the air. Now I want to show you how to turn that into a ground gradient, because that’s what actually matters for clearing obstacles on the ground. The key idea is a “wind factor.” Any gradient is the vertical distance divided by the horizontal distance. The air gradient is measured through the air mass; the ground gradient is measured over the ground. Wind changes the horizontal distance you cover over the ground in a given time, so it changes the gradient. Here’s the method. You take the TAS, divide it by the ground speed, and that ratio is the wind factor. Then you multiply the air gradient by that wind factor to get the ground gradient. Let me walk you through Example 1. An aeroplane has an air gradient of 12%, TAS 100 kt, headwind 20 kt. Now here’s the subtle bit — you don’t apply the full 20 kt headwind. You apply 50% of it. So 50% of 20 kt is 10 kt, and the ground speed becomes 90 kt (100 minus 10). Then 100 kt TAS divided by 90 kt GS gives a wind factor of 1.11. Multiply the 12% air gradient by 1.11, and you get a ground gradient of 13.32%. Notice what happened: a headwind makes the ground gradient steeper than the air gradient. That makes sense — you’re covering less horizontal ground per unit of height, so the slope over the ground is steeper. Example 2: same 12% air gradient, but now TAS 160 kt with a 20 kt headwind. Again, 50% of 20 kt is 10 kt, so ground speed is 150 kt. 160 divided by 150 gives a wind factor of 1.07. Multiply 12% by 1.07, and you get 12.8%. Now the tailwind case — Example 3. Same 12% air gradient, TAS 100 kt, but a 20 kt tailwind. Here’s where the asymmetry comes in. For a tailwind you apply 150% of the wind, not 50%. So 150% of 20 kt is 30 kt, and the ground speed becomes 130 kt (100 plus 30). Then 100 kt TAS divided by 130 kt GS gives a wind factor of 0.77. Multiply 12% by 0.77, and you get a ground gradient of 9.24%. Example 4: 12% air gradient, TAS 160 kt, tailwind 20 kt. 150% of 20 kt is 30 kt, so ground speed is 190 kt. 160 divided by 190 gives a wind factor of 0.84. Multiply 12% by 0.84, and you get 10.1%. So the tailwind flattens the ground gradient — you cover more horizontal ground per unit of height, so the slope is shallower. Now, that 50% headwind and 150% tailwind rule — I want you to remember it as a hard rule, because it’s not just a convenience. The note in the material is explicit: if the ground gradient is to be used for the calculation of obstacle clearance, the application of headwinds and tailwinds must include the 50% headwind and 150% tailwind rule. That’s a regulatory requirement baked into the performance calculations, so you don’t get to use the full wind component. Now let’s move to a full worked example — Example 5. This is a typical climb gradient question. We need to determine the ground distance for a Class B aeroplane to reach a height of 2000 ft above Reference Zero. The conditions: OAT 25°C, pressure altitude 1000 ft, gradient 9.4%, speed 100 KIAS, wind component 15 kt headwind. First, what is Reference Zero? Reference Zero is the point on the runway or clearway plane at the end of the Take-off Distance Required — the TODR. It’s the reference point for locating the start point of the take-off flight path. So when the question says “2000 ft above Reference Zero,” it means 2000 ft above that point at the end of the take-off distance required, not above the runway threshold or anywhere else. That’s the datum from which the take-off flight path is measured. So the whole picture is: you have the air gradient, you correct it for wind using the wind factor with the 50% headwind rule, and you get the ground gradient. Then you use that ground gradient to work out how much ground distance you need to reach a given height above Reference Zero — which is what you need for obstacle clearance. Let me show you the geometry. The air gradient is the slope through the air mass; the ground gradient is the slope over the ground. With a headwind, the ground gradient steepens; with a tailwind, it flattens. And the wind factor is simply TAS divided by GS. One thing I want to be clear on: the wind factor is always TAS divided by GS. For a headwind, GS is less than TAS, so the factor is greater than 1, and the ground gradient is steeper. For a tailwind, GS is greater than TAS, so the factor is less than 1, and the ground gradient is shallower. That’s the whole logic in one sentence. Now, in Example 5, we’ve got a 15 kt headwind. Applying the 50% rule, that’s 7.5 kt off the speed. But here’s the catch — the speed given is 100 KIAS, indicated airspeed, not TAS. So before you can compute the wind factor, you’d need to convert that KIAS to TAS using the OAT of 25°C and the pressure altitude of 1000 ft. That conversion is the step that ties the performance data together — the gradient of 9.4% is given, and you’d apply the wind factor to it to get the ground gradient, then divide the 2000 ft height by that ground gradient to get the ground distance. I want to stop here and make sure the method is solid in your mind, because this is the core of climb performance for obstacle clearance. Air gradient, corrected by a wind factor of TAS over GS, with the 50% headwind and 150% tailwind rule applied to the wind component. Headwind steepens, tailwind flattens. And Reference Zero is the point at the end of the take-off distance required — the datum for the take-off flight path.