# Tag Info

115

Your flight took off during a storm. During a storm, the wind speed close to the surface of the earth is much lower than the wind speed a bit higher up. This variation of wind speed over a short vertical distance is called wind shear. The aircraft is taking off into the wind, so during the initial climb the headwind increases. Increasing headwind during ...

56

It depends on exactly how you define "lift" and "weight". You might say intuitively that lift is all the forces acting on the aircraft in the upward direction, like this: In this case, lift must equal weight, otherwise the aircraft would be accelerating. That is, it's rate of climb would be changing. But it's more usual to define lift this way: Here, lift ...

21

The brief period of leveling off is not unique to the flight on that day. Looking at track logs for previous days, it always levels off at around 7,000 feet for some period. As Terry commented, this is most likely to deal with the congested airspace. Other routes coming into and out of the London area probably pass through the same space, and keeping flights ...

21

According to the performance data in the manual it takes 19.9 minutes to get to 70,600 feet. Refueling time will depend on how much fuel is needed for the mission as well as how long it takes to hook up to the tanker. The manual covers the operations for that but I dont see a flow rate of fuel in there. A lot of that stuff is covered in this podcast.

20

As the answers to your original question already explained, you do need extra lift to accelerate upwards. Once the wing is set into a vertical motion, however, lift again exactly equals weight to keep the wing at a constant vertical speed (if we neglect thrust and drag for a moment). No extra lift is needed to maintain that vertical speed. Only when you want ...

19

In an aircraft that is climbing at a constant vertical velocity, the total of the upward-directed vertical forces is the same as the total of the downward-directed vertical forces. Were it not so, the vertical velocity would not be constant, since any non-zero balance of the vertical forces would result in an acceleration...

16

What you say is true only for turbojets and aircraft with fixed-pitch propellers. Generally, all optimum points for variable-pitch propeller driven aircraft are at lower speeds than those of jet aircraft. The reason is the variation of thrust with speed: For propellers, thrust is inverse to speed, while it is roughly constant over speed for turbojet aircraft ...

15

Short answer: No. Long answer: When the flight path is not horizontal, lift will not be vertical but perpendicular to the direction of motion (in still air). Thrust will also have a vertical component and is different in magnitude from drag, because excess thrust is needed to increase the potential energy of the plane. Note that the vertical component of ...

14

Because ATC expects you to climb with at least 500 fpm when crossing to the next level. Otherwise you will block intermediate levels for an extended period of time. If you are climbing 2000ft to the next available level for your direction of travel, that will take at most 4 minutes at 500fpm or more. This means the flight level you are crossing will need to ...

14

Sometimes a departing airplane will receive or be offered an accelerated climb profile from ATC which gets it up and out of the published approach and departure patterns quickly. If the pilot elects to take the profile, the resulting climb will be unusual, and spectacular. I was on a Rockwell SabreLiner departing from San Jose, Ca in a very lightly-loaded ...

13

Short answer: Yes. But only for a moment. As you suggest, let's focus on the 90° case, because many things will become simpler. Intermediate climb angles can be covered by a linear superposition of horizontal and vertical flight. Now we shoot up at 90° flight path angle (engine thrust permitting) and leave some angle of attack for the wing to create lift. ...

13

The climb gradient is the percentage of the rise over run (100% if you are climbing at 45 degrees) that your aircraft is climbing at while the rate of climb is the speed at which you are climbing based off the airspeed and climb gradient (given in feet per minute).

12

Looks like an intermediate altitude assignment from atc. A look at the standard instrument departure plates (SID's) for Stansted shows that the final altitude on some of the procedures is 6000 ft. They probably had to stay at that altitude until the departure controller passed them off to the en route controller who assigned them their cruise altitude. On ...

12

If the pilot commands a certain glide path angle, any change in airspeed will also change vertical speed, because both are connected by the glide path angle $\gamma$. Note that the vertical speed is air speed$\cdot\sin\gamma$. Climbing or descending makes no difference; only the sign of $\gamma$ changes (positive $\gamma$ means the aircraft climbs). Also ...

12

Does the pilot flying have to manually "push the button" on the control panel? It's automatic, with a manual changeover button available to the pilots as shown below (label no. 2). How does the aircraft know when the transition should happen? When the target climb indicated airspeed (IAS) equals the target climb Mach number, the changeover takes place, ...

10

When you say, No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift. you're missing where the energy of the wing comes from. Lift isn't a magical power that creates potential energy out of nothing: it just turns airspeed (kinetic energy) into height (potential energy). In your example, the ...

10

The maximum angle of climb for all aircraft is achieved when the specific excess thrust available is maximum. $sin \ \gamma_{max} \ = \frac{(T-D)_{max}}{W}$ However, the thrust varies differently with speed in case of propeller and jet engines. Source: code7700.com For turbojet aircraft, the thrust is approximately constant with speed. So, $(T-D)_{max}$ (...

10

A climb gradient is a geometry problem -- the relationship of two points in 3-dimensional space... to get from one to the other you gain X' in Y NM, so you have X/Y feet per nautical mile as a climb gradient. Or it can be expressed in percent slope -- the two terms are different ways of describing the same thing. A climb rate is simply speed in the ...

9

The gradient of climb is the ratio of the increase of altitude to horizontal distance through the air, not over the ground. The definition used by the UK CAA in CAP 698 is: Climb Gradient The ratio, in the same units of measurement, expressed as a percentage, as obtained from the formula: - \text{Gradient} = \frac{\text{Change in Height}}{\...

9

Aircraft drag scales with the power of speed. This by itself should make clear that deviating from the point of optimum performance will always incur higher losses which cannot be made up for later. Let P₁ and P₂ be polar points away from the optimum point $\text{P}_{opt}$ along a quadratic drag polar. No matter how you interpolate between the two points (...

9

For the climb profile, the first two values are indicated airspeed -- below 10,000 and above 10,000 feet MSL, and the final value is a Mach fraction. So 250/280/78 would be a climb of 250 knots below 10,000', 280 knots above 10,000', and transition to a climb of Mach .78 somewhere in the high 20's or low 30's (i.e. at the point where 280 KIAS = M 0.78). For ...

8

Ron nailed it. (See comment on OP.) It depends on what you’re trying to achieve. Climbing “fast” in terms of maximum vertical speed = Vy (best rate) If the noise sensitive zone is immediately surrounding the airport, you want to use Vy to put altitude under you as quickly as possible. Climbing “fast” in terms of maximum feet per nautical mile = Vx (best ...

8

To enter the climb, it's power, attitude, trim. That is, start by setting climb power, as you already know. Remember to compensate for adverse yaw with the rudder. Immediately raise the nose to the climb attitude. Your instructor will teach you to recognise what this looks like in your aircraft: e.g. there might be a part of the dash that lines up with the ...

8

If we define lift as the component of the total aerodynamic forces on the aircraft that is perpendicular to its direction of motion, then lift will be slightly smaller in a stable climb. It is probably easiest to analyze the situation in a coordinate system that is tilted such that one of the axes is parallel to the direction of motion. Then all of the ...

7

I sort of feel that the rest of the answers are unnecessarily complex, given how simple the fundamentals here are: Momentum Question: Is it necessary that L>m.g (or as you put it, an excess of lift) in order to climb? Answer: No, at least not a sustained excess of lift. Newton's Laws state that an object in motion will remain in that state unless a force ...

7

Attitude + Power = Performance Performance is your airspeed, climb/descend rate, and rate of turn. Attitude is pitch, bank, and yaw. Power is RPM, if you have fixed pitch prop. So how do you transition from climb to level flight? When you reach 2000ft, or maybe 1950, push forward on the yoke or control stick to about level pitch, check the altimeter to ...

7

Here's Peter's diagram with two different airspeeds, so you can see the difference in the vertical speed:

7

If you keep increasing your angle of attack, your drag will increase and combined with the increasing along trajectory component (your path angle will increase) of the weight vector will slow you down (assuming your thrust does not exceed your weight + drag). Because you slow down, your lift will decrease. The lift will now be less than the across trajectory ...

7

So, 5,000 feet to gain and you're climbing at 750 ft/min. At that rate it would take 5000/750=6.67 minutes (to 2 decimal places) to make it. A speed of 90 MPH is 1.5 miles per minute. Thus 10 miles at that speed would be 10/1.5=6.67 minutes. Hmm, same value. I would suggest turning slightly to avoid the peak. Seriously, the difficulty is that you can't ...

7

It appears that it took the pilot about 15 mins to climb from near sea level to FL370. That's a climb rate of about 2500 feet/min, which is a pretty modest climb rate for a jet. If you're looking for fast climb rates in a transport, check out Lear 35s. If given an unrestricted climb to cruise altitude, they will often do 5000+ feet/min!

Only top voted, non community-wiki answers of a minimum length are eligible