General Principles - Climb — Page 195, Lesson 227

Let’s pick this up right where the climb segment begins, because that’s the heart of what we’re about to do. We’ve already got the air gradient — the climb performance of the aeroplane relative to the air it’s flying through. Now we need to convert that into a ground gradient, because the obstacle you must clear is fixed to the ground, not floating in the air. Here’s the key idea. The air gradient tells you how much height the aeroplane gains for a given horizontal distance through the air. But the wind moves that body of air over the ground. So the distance the aeroplane travels over the ground in the same time is different from the distance it travels through the air. That difference is exactly the wind component. Let me walk you through the worked example in Figure 3.38, because it shows the whole chain. We have a 15-knot headwind. The true airspeed, the TAS, is 104 knots. That TAS is calculated from the calibrated indicated airspeed, the KIAS, using your circular slide rule — at 1000 feet pressure altitude and 25°C, 100 KIAS equals 104 KTAS. So the aeroplane is moving through the air at 104 knots. Now, because the wind is a headwind of 15 knots, the ground speed is less. And here’s a critical point I want you to remember: wind speed is always expressed as a TAS, a true airspeed. So we subtract the wind from the TAS. 104 KTAS minus 7.5 knots gives 96.5 knots ground speed. Wait — why 7.5 and not 15? Because the climb segment is only half the wind. The wind component acting along the climb path is the full 15 knots only if you climb directly into it, but in this example the effective component is 7.5 knots. So the ground speed is 96.5 knots. Now we form the wind factor. TAS divided by GS — 104 divided by 96.5 — gives a wind factor of 1.08. That factor tells us how much more distance the aeroplane covers over the ground compared to through the air, per unit of time. Then we apply that factor to the air gradient. The air gradient here is 9.4%. Multiply 9.4% by the wind factor 1.08, and you get a ground gradient of approximately 10.15%. So the aeroplane, relative to the ground, climbs 10.15 units vertically for every 100 units of horizontal travel over the ground. Now, before we go further, I want to give you a piece of practical advice that I use every time: draw the question. Sketch the triangle, put in the known parameters — the TAS, the wind, the gradient — and the visual relationship becomes much easier to consider. You’ll see the geometry clearly. Now let’s move to Figure 3.39 and 3.40, because this is where we apply the gradient to a real obstacle clearance problem. We’re considering a Class B aeroplane. For a Class B aeroplane, the screen height is 50 feet. The climb segment begins at the screen height above Reference Zero. So if we want the aeroplane to be 2000 feet above Reference Zero, it only needs to gain an additional 1950 feet — because it already starts 50 feet up. Now, what does a 10.15% gradient actually mean? It means the aeroplane will be 10.15 units higher after 100 units of horizontal travel. So we need to find out how many times 10.15 divides into the required vertical height gain of 1950 feet. 1950 divided by 10.15 gives 192.12. That number tells us the vertical height gain is 192.12 times greater than the 10.15 units. So the horizontal distance will also be 192.12 times greater than the 100 units. Multiply 100 by 192.12, and you get the horizontal distance travelled in feet — 19,212 feet. Let me make sure that logic is crystal clear, because it’s the core of every climb gradient problem. The gradient is a ratio: vertical over horizontal. If you know the vertical height you need to gain, you divide that by the gradient’s vertical component to find the multiplier, then apply that same multiplier to the horizontal component. That gives you the horizontal distance required. Now let’s look at Example 6, which is a slightly different flavour of the same problem. A light twin-engine aeroplane has a 10% climb gradient after take-off. We need to find out by how much it will clear a 900-metre-high obstacle situated 9740 metres from the end of the Take-off Distance Available — the TODA. Remember the definition: percentage gradient is merely the vertical height for a horizontal distance travelled of 100 units. So a 10% gradient gives 10 units up for every 100 units along. That’s the ratio we work with. To find the height gain after covering a horizontal distance of 9740 metres, we set up the proportion. For every 100 units along, we gain 10 units up. So over 9740 metres, the height gain is 9740 divided by 100, times 10 — that’s 974 metres. The obstacle is 900 metres high, and the aeroplane gains 974 metres by the time it reaches the obstacle’s horizontal position. So it clears the obstacle by 974 minus 900, which is 74 metres. Now, one practical note before we finish. The height of the obstacle is above Reference Zero, and so is the start of the climb segment. That means we’re measuring everything from the same datum. And for practical purposes, a screen height of 50 feet is taken as 15 metres. So in this example, the climb segment starts 15 metres above Reference Zero, and the obstacle is 900 metres above Reference Zero — the height gain we calculated, 974 metres, is measured from that same Reference Zero. So the whole procedure is: convert air gradient to ground gradient using the wind factor, then apply that ground gradient as a ratio of vertical to horizontal to find either the height gained over a given distance, or the distance required for a given height gain. Draw the triangle, put in the numbers, and the relationship becomes obvious.