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

Let’s work through this one together, because it pulls together climb gradient, true airspeed, and the power relationship — and it’s a classic exam-style calculation. We start with an aircraft climbing at a 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. First, we need to convert IAS to TAS. Why? Because rate of climb is a vertical speed, and the climb gradient relates vertical distance to horizontal distance travelled through the air. That horizontal distance is the true airspeed — the actual speed of the aircraft through the air, not the indicated speed. At 8500 feet pressure altitude and 15°C, using a circular slide rule, 85 knots IAS converts to 100 knots TAS. So we have 100 KTAS. Now, a key simplification: at climb angles less than about 20 degrees — and in practice they always are — the difference between the hypotenuse and the adjacent side of the right-angled triangle is so small that we disregard it. So we treat the TAS as if it were the horizontal component, even though the aircraft is actually flying up the hypotenuse. EASA makes the same assumption, so your answers will be correct. Now, the gradient of 3.3% means that for every 100 units of horizontal distance, the aircraft climbs 3.3 units vertically. So we can think of the horizontal component as 100 and the vertical component as 3.3. For rate of climb, the horizontal component is the TAS — 100 knots. We need to convert that into feet per minute. 100 knots TAS multiplied by 6080 feet per nautical mile gives 608,000 feet per hour. Divide by 60 minutes per hour, and we get 10,133 feet per minute. That’s the horizontal speed in feet per minute. Now, since the gradient is 3.3 per 100, we divide 10,133 by 100 to get 101.33, then multiply by 3.3. That gives us 334 feet per minute. So the rate of climb is 334 ft/min — which is option d. Now let’s step back and look at the underlying formula. The gradient of climb is given by Thrust Available minus Thrust Required, divided by Weight. That’s (T - D) / W. This tells us the ratio of excess thrust to weight — essentially the climb angle. To get rate of climb, we multiply that gradient by the TAS. So Rate of Climb = (T - D) / W × TAS. But there’s more detail. The velocity here is True Airspeed, and 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. That’s the key insight: rate of climb is essentially excess power divided by weight. The excess thrust gives you the gradient, and multiplying by TAS converts that into a rate of climb — because power is force times speed. So in this problem, we used the gradient directly with the TAS converted to feet per minute, and we got 334 ft/min. That’s the answer.