Why does the faster-moving air over the static port not result in a lower static pressure?

I understand the concept that the static port is measuring "static" pressure, meaning the pressure the air is exerting on its surroundings. However, there is also Bernoulli's principle, which, put simply, states that as the velocity of air increases, the pressure it exerts on its surroundings decreases. We see this in effect in venturis, of course, but it seems to me that Bernoulli's principle would also imply that the pressure observed from the static port would similarly decrease as the aircraft accelerates. Obviously this doesn't happen however, since the altimeter is unaffected by acceleration.

Thus my question: why does the faster-moving air over the static port not lower the pressure it exerts?

Here is a video explanation of Bernoulli's principle being used in pitot-static systems, which seems to suggest the static port would indeed observe a pressure decrease (I've labeled the part I think the narrator is using as the static port):

• Why do you think the flow is accelerated over the static port compared to upstream?
– JZYL
Commented Dec 5, 2019 at 2:54
• There is a small effect but the designers try to minimize it when they place the ports on the aircraft. That’s why the altimeter reads high and there is a momentary jump in the VSI when you open the alternate static port inside the aircraft. Commented Dec 5, 2019 at 3:52
• Related, probably not a duplicate: Does the pressure at the static ports drop as the aircraft's speed increases? Commented Dec 5, 2019 at 4:36

Bernoulli's principle would also imply that the pressure observed from the static port would similarly decrease as the aircraft accelerates.

It's not that simple. Aircraft are complicated shapes. The fuselage is composed of all sorts of angles of attack, things sticking out into the airflow, and the pressure distribution of air across it is very uneven, just like that of a wing is. Take a look at this diagram from an FAA mechanics publication.

There are zones (like points 2,3, and 5 in the diagram) where the static pressure deviation is zero. This is where the engineers put the static ports.

Another way to think about this is that you're right that the airflow over the static ports should make the pressure lower--- but the question is: lower than what? The answer is: lower than the airflow's stagnation pressure, when the outside airflow is at rest (relative to the aircraft). Which happens at the nosecone, or the leading edge of the wing... or at the pitot tube. It's lower than that by the same amount that the pitot pressure is higher than ambient, making the pressure at the static port exactly ambient outside pressure. Which should make sense, because in the reference frame of the air, that static air isn't doing anything but sitting around waiting for an airplane to come by.

The Bernoulli effect does, however, have an overall average result: the cabin pressure in the airplane is lower than static, because the angle of attack of the various surfaces of the fuselage generates suction, pulling a small amount of air out of the cabin, just as it does for the wing if there are little holes in the wing. Which is why you get the little blip when you pull the Alternate Static open.

However, there is also Bernoulli's principle, which, put simply, states that as the velocity of air increases, the pressure it exerts on its surroundings decreases.

I think that's accurate.

So let's suppose you're flying at 120 knots. The air ahead of the aircraft is moving at 120 knots (in the aircraft's frame of reference). The air immediately outside the static port is also moving at 120 knots. The air behind the aircraft is, of course, also moving at 120 knots.

Since the speed of the air is the same in all three locations, according to Bernoulli's principle, the pressure should be the same in all three locations, too.

So, we've established that the air pressure immediately outside the static port is about the same as the air pressure several feet away from the aircraft. Can we conclude that the air pressure inside the static port is also the same? Bernoulli's principle doesn't tell us the answer. The main reason is that Bernoulli's principle only applies along a streamline, and the air outside and inside a static port are not on the same streamline.

But in any case, Bernoulli's principle doesn't give us any reason to think that the air pressure in a static port should be lower than in the surrounding air.

As pointed out by JScarry, the position and shape of the static port opening on the airframe are carefully chosen to minimize any ram air or suction effect and thereby sense the static pressure as accurately as possible.

In homebuilts, the static port location has to be determined by experiment, and if your experiments are sloppy, you'll find it possible to fly 250 MPH on a 65 HP engine (I'm exaggerating a bit here; I knew someone who built a kitplane with a homebrew static port and attributed its impossibly good top speed to his superior construction technique ;-))