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There is a big difference between flying a circuit to a parallel runway and joining a 3 mile final. If the reports are accurate, this was the equivalent of aircraft from opposite directions joining straight in approaches to parallel runways from the boundary of the class D at Moorabbin, i.e. over the beach at Aspendale. Would you typically be at 100 KIAS 3 miles from the airfield or a higher speed? If you are mixing with larger aircraft, keeping the speed up until closer in may be desirable.

According to the report I read the collision occurred 3 miles from the threshold, and the runways are 200m apart. My 1 in 60 calculation gives a margin of error of around 1 degree at that distance (or maybe 1/2 a degree, if you want to ensure some separation between the aircraft). If the Cirrus actually flew through the opposite centreline he was misaligned by 2 degrees. That seems pretty high precision to ask of a visual alignment. I'm not sure about tolerances for an ILS.

Carrying the Fire I agree - a very good book

Airspeed should be maintained at the target (recommended) approach speed. Slowing down to lose height is OK, until it isn't. Short field approaches are pretty much by definition riskier than normal approaches - otherwise you would always use the short field speed. Some aircraft have significant inaccuracies in the airspeed indication at slow speeds, so you might not have the margin you think you have.

The restrictor and return line is not there to adjust fuel pressure. If your fuel pressure is incorrect you should find the real reason why.

I wouldn't really recommend this. You certainly get more drag as you slow down but you are also getting closer to stall. You should be keeping speed stable at the target airspeed on approach. If idle power and flaps are not enough to fix the approach, either a slip or go-around are the best options. Depending on the aircraft, increasing speed might also be an option. A C152 at VFE with full flaps has an impressive angle of descent, and loses the speed again quickly. However the Jabiru I flew didn't have enough margin between VFE and approach speed to make this a useful technique.

There was a startup called Better Place trying to build a network of battery swap stations for EVs. It relied on standardized batteries accessible from under the car. In order to access the battery switch station, Better Place customers would have to swipe their membership card. The remaining process was fully automated, similar to going through a car wash, so the driver never had to leave the car The car owner wouldn't own the batteries, they just purchased the electricity. So all the problems with battery degradation etc. would be managed by Better Place across their whole inventory rather than being a risk to the car owner. https://en.wikipedia.org/wiki/Better_Place_(company) Hard to know whether its a better or worse idea than charging stations. Their biggest problem appears to have been no-one driving electric cars at that time.

A reasonable estimate is that a single pump can deliver enough fuel in 1 hour across multiple vehicles to drive 5000 km in total. How much power do you need to provide enough electricity in 1 hour to drive 5000 km?

I don't mean operating continuously. Check my math: My car can do about 800 highway km on a tank. Lets say you can fill 1 car every 10 minutes, that's 4800 km, call it 5000 km worth of energy dispensed per hour. A Tesla supposedly uses approximately 20 kw/h per 100km so 5000 / 100 * 20 = 1000 KWh or 1 MWh per hour = 1 MW equivalent. I think I estimated 10 cars/hour and a more efficient car when I did the original calculation, but that is the order of magnitude. I was prompted to think about it when we passed service stations on the Hume at Easter with queues of cars at every pump. Sure you can charge at home off peak most of the time, but it's the peak capacity when everyone wants to travel on a holiday weekend that poses the problem. Don't get me wrong, I want an electric car ASAP and I know we have to go that way, but we also need to understand the obstacles we have to overcome. I don't believe we are going to get there by waiting for the market to move. Car makers and buyers will wait until the infrastructure exists, and infrastructure builders will wait till the demand exists.

Are there any models without regenerative braking? Regenerative braking is the hybrid's main advantage - all the energy comes from the ICE engine, but regenerative braking allows them to capture much of the energy normal cars lose as heat in the brakes. My point was that the hybrid is probably worst case cost and complexity wise. You have ICE, transmission, fuel system, cooling system, plus electric motor and battery. And it's still only 2K over the ICE only model. There are a surprising number of Tesla's getting around these days. I know someone who regularly drives between Melbourne and Ballarat in their Tesla. We traveled up the Hume at Easter, and were passed by several Teslas on the way. The technology is there when the large car makers decide they want to do it. What we do not have is the charging infrastructure for mass adoption. I think we underestimate the capacity of the energy distribution infrastructure we have built around internal combustion engines. I was doing some back of the envelope calculations, and estimated that a single petrol pump delivers the energy equivalent of about a 2 megawatt charger. So a 10 pump petrol station might be the equivalent of 20 MW charging capacity. And there's another one across the road, and another around the corner. One fuel tanker delivering petrol might have equivalent energy to the South Australian grid battery.

Yes, looks like fun for local flying. The biggest problem if you have limited energy is drag, so I think anything practical to go further will look more like a motor glider i.e. very streamlined with a big wingspan to minimize induced drag. Pipistrel seem to be going that route, their electric aircraft are interesting. If they put their electric powerplant in their Sinus airframe it seems like it would be one of the most practical electric options.

I think Tesla and Toyota at opposite ends of the spectrum show that many EV problems are largely solved. Tesla has range, fast charging and power (see ludicrous mode). Toyota put a battery, electric motor, regenerative braking etc into their hybrid models and only charge a couple of thousand extra on top of the petrol model. If Toyota want to build a full electric car they need a bigger battery and electric motor, but the cost of that should be more than made up by being able to delete the ICE, fuel & exhaust systems, most of the transmission etc. The reason electric cars are expensive is that we don't have the charging infrastructure for mass adoption, and (a) if you can't sell to the masses you have to recoup development costs over a smaller number of vehicles and (b) if you can only sell to a few percent of the population, you might as well target the rich people. Aircraft are a different problem. Cars sit on the ground, they don't need energy to hold them up. With an aircraft, every kg requires energy to keep it in the air. As aircraft get bigger you quickly get into diminishing returns. I don't think that improvements in battery technology will ever solve that problem - there are limits to the energy available from the chemistry that can't be worked around. Synthetic liquid fuels for aviation use seem much more likely to me. https://thebulletin.org/2009/01/the-limits-of-energy-storage-technology/

I don't think anyone is arguing with the idea that they refer to different things - rather that it's so obvious it goes without saying. The argument is whether something with a mass of 1 kg weighs anything other than 1 kg, when "weigh" is used in it's everyday usage i.e. stationary on the surface of the earth, ignoring local variations in gravitational force, and using kg to refer to kg-weight. From Wikipedia: The kilogram was originally defined in 1795 as the mass of one litre of water. This was a simple definition, but difficult to use in practice. By the latest definitions of the unit, however, this relationship still has an accuracy of 30 ppm. i.e. the difference between the mass of 1 litre of water and the current SI definition of a kg which uses the Planck constant is less than 0.003%. For our purposes, the mass of 1 litre of water is close enough and much simpler to understand. If you take 1 litre of water, which has a mass of 0.999972 kg using the SI definition and put it on your scales, it is going to read 1 kg within the limits of accuracy of any regular set of scales. All this reference to the Planck constant, Avogadro's number etc. is just obfuscation - it is not necessary for discussion of the topic.

You just defined force yourself, with the qualifier "when unopposed". "When unopposed" is the important bit. F = m.a BUT F is the vector sum of all forces acting on the mass. We can only calculate a force using F= m.a if there is only one force, or all other forces are known. It is very common e.g. for there to be 2 forces perfectly opposed, and therefore acceleration is zero, net force is zero, but significant and measurable force is being applied. An example is standing on a set of scales. The scales measure a force, but there is no change in velocity. If I use a spring compressor and apply a force of 2000 newtons and compress a spring by 50cm, I have applied a force. I have done work, as evidenced by the energy stored in the spring. Where is the mass? Where is the acceleration? We know the value of F, but there are no values for m or a to use in the equation. If the compressor suddenly breaks, we will have mass (the spring and parts of the compressor) and acceleration. F = m.a suddenly applies. Force is the potential to accelerate a mass when unopposed (as you said in your first definition) which is why the units of measurement incorporate mass and acceleration. It does not mean that you can't have force without acceleration - which is what defining force as F = m.a implies: if a = 0, F = 0. To repeat myself, to apply F = m.a, F must be the sum of all forces. The equation doesn't tell you the value of any specific force if there is more than one.

We do not agree. That is an equation about acceleration. You can have a force where acceleration is zero. If you push a box of bricks across the floor at a constant speed, you are exerting a force to overcome friction. You can measure the work done as force x the distance you moved the box. Acceleration (once the box is moving at a constant speed) is zero, but you have a force that must be used if you want to calculate e.g. work and power. Likewise, when an aircraft is straight and level at a constant speed (i.e. no acceleration) we do not say that there is no force acting on the aircraft. If we add all the forces we get zero net force, true, but that does not help us define force. You do like complex explanations. That doesn't really answer "what is a kilogram?", it answers how much is a kilogram. Previously there was a lump of stuff with a mass of exactly 1 kg for comparison, but the problem is any physical object can gain or lose mass to the environment. So that is the answer they came up with for how do you define a kilogram without using something that can physically vary. For our purposes, the mass of 1 litre of water, or the mass of a reference object of 1 kg is almost certainly sufficient. A kilogram is the measurement of mass: how much "matter" is in an object. Effectively a measure of how many protons, neutrons etc. it contains in it's atoms. (Can we ignore relativity please!) Kilogram-weight is a measure of force - the amount of force exerted by gravity on a mass of 1kg on earth. We want to measure kilograms (mass), but scales measure force. Calibrating them in kilogram-weight and calling it kilograms is convenient. We could measure newtons and divide by 9.8, that would be less convenient.

That diagram is a turn not a roll. The force vectors are totally different. Nor is is a diagram of the result where the AOA remains unchanged.

This video shows some interesting instruction on rolls: The whole video is worth watching, but there is discussion on the slow roll at 17:18 and hesitation rolls at 24:30. The description of the hesitation roll is that it is the same as the slow roll, but with the hesitation added. This really dispels the idea that the pilot rolls the aircraft and then corrects the flight path. It is simultaneous aileron, elevator and rudder inputs.

I said lift required was infinite for a turn with 90 degrees of bank. Of course you are right. Lift required trends to infinity as you approach 90 degrees, but at 90 degrees you need to divide by zero: infinity is not the answer, there is no valid answer.

Maybe if you post the answers instead we will have more luck trying to work out the question you want to ask. You introduced the question with a video, that suggests that you are intending to post a question that relates to real aircraft behaviour. Incidentally, the video includes real time G readings which actually show us the lift the wing is producing at each point. Not surprisingly they differ from the "perfect" roll I described. There are limits to how perfectly a pilot can fly, and also limits to what an aircraft is capable of.

That is true if the aircraft is instantaneously rotated around an axis perfectly aligned with the airflow. But it's not real life. In real life, even at maximum roll rate it takes time to go from 0 to 45 degrees. It also depends on the skill of the pilot, but lets assume the pilot can fly the roll perfectly. As the pilot begins the roll the lift vector is pointed to the side, which would begin to turn the aircraft. To prevent the turn, opposite rudder is required and the aircraft is in uncoordinated flight. At 45 degrees, the wing provides half the lift, and the side force on the fuselage due to uncoordinated flight provides the other half. The wing and fuselage produce equal and opposite sideways forces which means the aircraft does not turn. The wing is providing half the vertical lift at a 45 degree angle. That requires less total lift than supporting the whole aircraft at zero degrees bank. To fly it perfectly, the rudder needs to be increased and AOA of the wing decreased as the aircraft rolls from 0 to 45 degrees. That is for a 45 degree coordinated turn, not a roll. A turn is totally different to a roll. In a turn, lift required increases as bank angle increases. In a roll, lift from the wing decreases as bank angle increases. At 90 degrees angle of bank in a roll (knife edge flight) the lift required from the wing is zero. At 90 degrees angle of bank in a turn the lift required from the wing is infinite i.e. a coordinated 90 degree banked turn is impossible.

An AOA indicator that according to the expert you quote "can’t tell AoA" doesn't work.

APenNameAndThatA pointed it out and how many posts did you spend arguing? I knew what you meant and it didn't seem worth worrying about.

Rather than bringing other planets into it, maybe you should re-read what you wrote. You made a simple unit conversion error - switching from kg to newtons without a conversion. 3 newtons should have been 30 newtons, rounding 9.81 to 10.

Someone might have been looking for a higher level of proof than OME declaring himself correct. I think you have misunderstood what Mr Munn wrote - selectively picking out bits that support you. From your post of his reply: We can and do derive AOA much more accurately and reliably from the airspeed indicator. A stopped clock is also correct for a given set of parameters, but like the spirit level AOA you need enough additional references to know when that is that the original reading becomes redundant. That is correct. It displays attitude not AOA, but only in one axis and only in unaccelerated flight. The attitude indicator shows the same and more information more reliably e.g. it isn't affected by acceleration.

There's a chance to earn $500 for your favorite charity then. (The offer only applies once i.e. the first person)