The pressure altitude is 8500 feet, and the outside air temperature is 15°C — Page 202, Lesson 233

Let’s start with the problem itself, because it pulls together everything we’ve been building. We have an aircraft with a climb gradient of 3.3%, flying at an indicated airspeed of 85 knots. The pressure altitude is 8500 feet, and the outside air temperature is 15°C. The question asks for the rate of climb, in feet per minute, and gives four options: 284, 623, 1117, or 334 ft/min. Now, the key idea here is that rate of climb is a vertical speed, but our gradient is a ratio of vertical to horizontal. So we need to convert that horizontal component — the true airspeed — into a vertical rate using the gradient. First, we have to convert IAS to TAS. IAS is indicated airspeed, what the instruments show. TAS is true airspeed, the actual speed of the aircraft through the air. At 8500 feet pressure altitude and 15°C, using a circular slide rule, the TAS comes out to 100 knots true airspeed. That’s the horizontal component we need. Now, there’s a simplification here that’s worth noting. At climb angles less than about 20 degrees — and in practice they always are — the difference between the hypotenuse and the adjacent side of a right-angled triangle is so small that we disregard it. So we don’t worry about the fact that the aircraft’s TAS is actually along the hypotenuse. EASA makes the same assumption, so your answers will be correct. So, with a gradient of 3.3%, we can think of it as a horizontal component of 100 and a vertical component of 3.3. That’s the ratio of “up” to “along.” For rate of climb, the horizontal component is the TAS, which is 100 knots true airspeed. We need to convert that into feet per minute. 100 knots true airspeed, multiplied by 6080 feet per nautical mile, divided by 60 minutes per hour, gives us 10,133 feet per minute. That’s the horizontal speed in feet per minute. Now, to get the vertical rate, we divide that by 100 — because our gradient is per 100 units of horizontal — and then multiply by the vertical component, 3.3. So 10,133 divided by 100 is 101.33, and 101.33 times 3.3 gives us 334 feet per minute. That’s the rate of climb. So the correct answer is 334 ft/min. Now, let’s step back and look at the formula behind this. The gradient of climb is given by Thrust Available minus Thrust Required, divided by Weight. That’s the formula: (T - D) / W, where T is thrust available, D is thrust required, and W is weight. That gives us the gradient. To get rate of climb, we simply multiply that gradient by the true airspeed. So rate of climb equals (T - D) / W, multiplied by TAS. But there’s a subtlety here. The velocity is true airspeed, but thrust and drag are forces. Force multiplied by distance gives work, and work divided by time gives power. So instead of thrust multiplied by velocity, we now have Power Available. And instead of thrust required multiplied by velocity, we have Power Required. So the rate of climb formula becomes Power Available minus Power Required, divided by Weight. That’s the deeper understanding — rate of climb is essentially the excess power, divided by weight. Let me make sure that’s clear. Power is the rate of doing work. Thrust times velocity gives power. So when we multiply the gradient formula by TAS, we’re converting the force-based gradient into a power-based rate of climb. That’s why the formula for rate of climb uses Power Available and Power Required instead of just thrust and drag. So, to recap: we converted IAS to TAS, used the gradient to find the vertical component, converted TAS to feet per minute, and got 334 ft/min. And the underlying formula is rate of climb equals (Power Available minus Power Required) divided by Weight. That’s the full picture. Does that make sense so far?