Recent content by Arjan82

That took me a bit 😆

This is way to complex to solve in a few forum posts. You need a physics book about turbulent jets for this.

So I agree, the gravitational potential energy is the same as long as you take the same reference level. I was confused because normally it makes no sense to talk about gravitational potential energy from a wing-fixed reference frame. I also agree that in an earth fixed reference frame the...

I agree There is not zero energy change due to a wing. What I'm saying is that from the perspective of a reference frame attached to the wing, the wing does not add energy to the flow. But if you want a complete picture, you now also need to include the earth. Because now it is the earth that...

Depends entirely on the setup as I said earlier. You have a pressure gradient, so it depends where on this gradient you put the static port.

You're misunderstanding me, I'm also not talking about the pressure gradient vs altitude. But any object that comes towards you through the air wil have an increased pressure in front of it that is needed to push the air away that is in front of it. That's the pressure I was talking about. This...

Yes. This is because for the air to accelerate, you need a pressure gradient. A pressure gradient is how you apply a force to an air parcel. And to get a pressure gradient the pressure needs to be higher in front of the rocket than behind it.

Both are and both are not. It depends, that's the whole point. So to repeat: Case 1: An object that accelerates in stationary air, so an accelerating rocket in free flight. In this case the static port will always measure the same static speed which is equal to the surrounding pressure (if...

But in that case Bernoulli doesn't apply anyway. In the whole discussion we (at least I am) talking about potential flow.

I can save you a lot of trouble. Look again at this equation: You can easily rewrite this equation to be expressed as an expression for the static pressure: $$ p = \rho \left(f(t) - \frac{\partial \phi}{\partial t} - \frac{1}{2}\left|\nabla \phi\right|^2 \right) $$ If you assume acceleration...

That is indeed what you also stated, and that is true, I'm not arguing that. But what you also stated is that if you are in the frame of reference attached to the wing, energy is added to the flow. And that is NOT true. You need to keep your eyes on the ball, because you are moving the...

Please re-read the appropriate books on this topic. Nobody is denying that a glider is converting potential energy into kinetic energy of the air. But that's only true from the earth-fixed reference frame. The question is whether you can apply Bernoulli on a wing. I say yes, because no energy...

Depends on what you want to do with the results. It is not really correct. It depends on the next steps.

Since your original question was about what a well-placed static port would measure on an accelerating rocket, I would assume you mean a rocket in free flight. So now you are saying you are talking about a rocket that is bolted on the ground. So I'm lost to what your question now actually is...

In real life nothing is instantaneous.