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The important V speeds Vs0, Vs1, VFE are IAS not TAS. Maneuvering speed is IAS but reduces with weight. The only speed that might be TAS is VNE.

It is definitely valid to decrease weight for an airport that is hot or high. GA aircraft will normally have performance charts that allow you to figure out whether you can take off or land in the space available at a specific weight. If the answer is no, one solution may be to reduce weight. The same applies to any aircraft, however it is more difficult if there are no performance charts for weight. It is still good to be aware of the problem. You can't use the same logic to increase weight because you don't know the limiting factors for MTOW.

Also the engine power is reduced (unless you have a turbo) so acceleration is reduced which increases the time it takes to get to the higher speed.

I'm not sure what you mean by "flys badly". If it truly flys badly it probably isn't a temperature thing. There are a number of factors that might influence MTOW: - Structure - how strong is the airframe? - Regulations e.g. a limit on MTOW or stall speed - Performance - the ability to climb after takeoff, or go around in landing configuration Temperature and altitude (i.e. air density) affect the true airspeed you need to fly and also the power available from the engine i.e. decreasing density has a double effect. It is possible to translate a high temperature into an equivalent lower temperature at a higher altitude. This is part of the performance subject in the PPL syllabus. It's more likely to be part of runway distance calculations than absolute MTOW though. I guess it is possible to have density limits on MTOW for climb performance, but I can't think of any examples.

I don't think coastal was specifically offered, only implied. The report says no tracking information was given, only that a clearance would be available "at or below 1000 feet". Coastal would have been 10 miles left of track, whereas 5 miles right of track would have taken him around the airspace at his 6500 cruising level. He seemed to be going right of track and around the airspace before being offered at or below 1000 feet.

He was denied a clearance, and turned right to avoid the CTA. Then he received a clearance from the Class D controller with the instruction "not above 1000 feet" and turned back on track and descended. The only problem was the ground level on track was 3000 feet.

You specifically said making a turn. A change in the mass due to fuel burn is one of the changes it is simpler to ignore. But the force resulting in THAT momentum change is drag.

The source of the force that makes the change is the wing. We bank the aircraft, which means that there is a sideways component to the lift force. This sideways force turns the aircraft.

Lets go back to the original statement and look at it in more detail: OK The laws of physics don't allow it. It's like dividing by zero in the middle of a sequence of calculations. Everything that follows is invalid. After a turn, if kinetic energy is unchanged, airspeed must still be 150kt. If momentum is unchanged you are still travelling at 150kt in the original direction while pointing 90 degrees to the original path. (In a helicopter maybe?) It is important that people understand this is wrong. Because airspeed, velocity, momentum, lift are all relative to the air, not the ground.

A turn is a change of momentum. The change of momentum is what is being analyzed. You can ignore some things e.g. kinetic energy and even gravity and get a useful answer. But not momentum.

The speed and quantity of kinetic energy do not change. Momentum definitely does change. Ignoring that is ignoring the laws of physics. The question is fundamentally about the force required to make the change - ignoring it makes no sense.

Velocity is a vector which means if you change direction, velocity and momentum have by definition changed. The force to change comes from the lift of the wings acting towards the centre of the turn.

#4 is wrong so any conclusions are invalid. Momentum has a direction. Momentum is not conserved, it is changed from one direction to another. That requires a force - the lift produced by acting on the air. Forces are acting on the air, not the ground so everything needs to be worked out relative to the air mass. The ground is irrelevant, apart from visual illusions.

A standard circuit is pretty clearly defined in AIP. If you're looking for a particular person, I don't know. I suspect it predates WWII.

CASA is the obvious answer. The main purpose of flying a circuit is to be predictable, and to observe and be observed by other traffic. On base leg, you you should be looking for traffic flying a straight in approach, and also possible other traffic that might have turned base from e.g. a closer downwind. Many aircraft have blind spots in a turn that would make that difficult. Straight legs of the circuit allow you to see traffic that is obscured in the turn. Additionally, AIP calls for a straight final approach beginning at least 500' AGL. The purpose is to allow you to check for traffic on the runway, and to make a stable approach. My understanding of the military initial and pitch is that it helps fast aircraft slow down, and creates separation for landing between multiple aircraft in a formation. It does not seem to be a good solution where you need to fit in with other unrelated aircraft in the circuit.

Performance calculations are often done before takeoff, but it may also be necessary to do them in flight. You don't use the lift equation in the air, but you learn it on the ground so that hopefully in the air you understand the effect of air density, the limitations when you are doing 120 knots in an aircraft with 50 kt stall speed etc.

It is important to understand the effect of air density on lift. It is a critical part of performance calculations. It is also important to understand the effect of velocity - that lift available is proportional to the square of the velocity. Other than that you have wing area (a bigger wing can lift more - not complicated) and the coefficient of lift. The coefficient of lift is just a constant that allows you to tie all the other numbers together allowing for different characteristics of different wings. As for the spirit level: Imagine you are in an aircraft climbing at a 5 degree angle at 60 knots. Now you close the throttle and glide, and the aircraft descends at a 5 degree angle at 60 knots. The angle of attack remains the same if the airspeed and weight are the same. There is virtually no change in the angle of attack from the climb to the descent, but the angle of the fuselage has changed by 10 degrees. The reading on the spirit level will change by 10 degrees - more than half the typical stall AOA - but the real angle of attack has not changed. If it was possible to read angle of attack with a spirit level you can be sure it would have been done long ago - but it is not.

I have been re-reading my info on switches. Switches can have different ratings for resistive loads vs. tungsten loads however the switches that are readily available to us typically seem to list one value. Which value is it? Are we asking for problems using e.g. a 15A switch for a 12V 100W light? Re wiring: - I know the capacitor is not connected to the battery with the master off - even though it remains charged, it can be discharged. - I know the battery is disconnected from the bus and from the regulator with the master off. - I know the regulator is disconnected from the bus with the master off - I know the regulator C terminal is disconnected from the battery with the master off - I think the capacitor remains connected to the regulator B terminal when the master is off - The capacitor does not discharge, so it isn't connected to the C terminal with the master off. The C terminal is the one I am not sure about. The requirements are difficult to satisfy: - needs to read the battery voltage within 0.2V - needs to be disconnected from the battery at shutdown - must not be disconnected from the B terminal while the engine is running or regulator damage can occur The last one is the one that might be a problem for me. I think the C terminal is connected to the bus which means it would be disconnected if I shut off the master. I don't think previous versions of the installation manual were as explicit. I could connect it to the B terminal, but then the voltage would be less accurate relative to the battery because it would be influenced by the electrical load on the alternator. Ideally perhaps I should run both B and C wires back to the battery, and another separate wire forward to the bus. Then they all need some sort of fuse at the battery end...

Sure, but when there are a lot of assumptions it's also worth verifying that the engineering calculations successfully predict what you see in the field. If a tungsten lamp inrush is similar, should we have a soft start for a light switch? It also depends on the wiring - I know in my aircraft the capacitor remains charged when the master is switched off so there won't normally be a large current when it is switched on. I can't remember the details of the switching of the charging system - I will have to double check what is switched.

How likely is that to be a real problem? I have never heard of it in practice. That switch needs to be rated to open a 20A+ DC circuit - is the capacitor going to draw enough energy to damage the switch when it closes?

The installation manual shows 20A at 5000 rpm, and lists the maximum output as 22A. Also, as you point out the capacitor can draw a much higher current from the battery for an instant when it is initially connected. You obviously want to be sure that the fuse won't blow unless there is a real problem, otherwise it becomes a reliability issue in itself.

The diagram is incomplete... we need to see the C regulator terminal and switching to understand the design. Rotax call for a 25A slow blowing fuse to protect the regulator. Since that circuit needs to take the entire regulator output anyway, the fuse between the regulator and capacitor seems redundant. The fuse on the battery side protects that circuit. I'm not convinced that a 25A fuse would provide much protection for the capacitor. My understanding of the capacitor function is: a) the PM alternator output is not very smooth because it comes from a single phase, and the capacitor helps there. Sure, a better alternator might be nice, it might also be heavier, bulkier, more expensive. b) if the battery and load is disconnected, it slows the voltage rise enough that the regulator should be able to respond and cut the output without damage to the regulator & charging circuit.

There are a lot of square inches of wing. The pressure difference required works out to around 0.1 psi for a typical light aircraft.

Would you be able to turn to starboard if the rudder was attached to the extreme port side of the boat? Of course, you would, so it’s not drag doing the job.

No, it's fundamentally wrong. I don't even know which side you are suggesting the lift comes from. Dividing people into Bernoulli and Newton camps is completely misunderstanding what Bernoulli and Newton are about. They are both correct* - they are just different ways of looking at the same thing. It depends on whether you choose to view air as a fluid or a stream of molecules. * Some people - usually those who divide people into Bernoulli and Newton camps - use Bernoulli's principle incorrectly. But that doesn't mean Bernoulli was incorrect.