30
$\begingroup$

I remember reading a book about private piloting techniques, (I can't remember the title; it has been so long; perhaps I can check my library history or something to see later), and I saw something bizarre.

The author was discussing instruments, and what to do in case of failures. He wrote something about how, in the case of a failure of a critical instrument, in a situation where you really need it, that one should punch out (or otherwise destroy) the glass of the gauge, to allow air into the system.

What?

Is there any way something like this could be useful, or at least theoretically logical? Can anyone deduce any sense out of this?

Improve this question
$\endgroup$
4
  • 5
    $\begingroup$ Ah, the old percussive maintenance. $\endgroup$
    – dotancohen
    Oct 21, 2014 at 14:23
  • 1
    $\begingroup$ Yes, it's a thing, but many airplanes today have an alternate static source that can be switched on instead... $\endgroup$
    – Brian Knoblauch
    Oct 21, 2014 at 16:38
  • 3
    $\begingroup$ Once while I was flying a C172 or PA28 there was a gauge I was suspicious of the reading. When I tapped the face of the gauge with my fingers a couple times, the needle set itself to about where I thought it should be. I believe it was the oil pressure gauge and after the tap it matched the (illuminated) low oil pressure annunciator. $\endgroup$
    – Ed Griebel
    May 19, 2015 at 17:53
  • 1
    $\begingroup$ Low oil pressure? Not good! $\endgroup$
    – Bassinator
    May 19, 2015 at 20:43

2 Answers 2

Reset to default
35
$\begingroup$

If the static port gets blocked in an aircraft with no alternate static port, you can break the glass of the VSI (Vertical Speed Indicator) to allow air from inside the cockpit in the static system. By doing this you will sacrifice the VSI in favour of the altimeter and airspeed indicator

Due to venturi effects, air inside the cockpit has usually a slightly lower pressure than the pressure measured by the static port outside your aircraft.

This causes your altimeter to indicate a higher altitude and the airspeed indicator will also indicate a greater speed. The VSI is useless after you break it.

Needless to say this does not apply to pressurized cabins nor to glass cockpits...

This works because the inside of the VSI contains a diaphragm in a pressure chamber. The diaphragm is connected to the static system through an unrestricted tube. The chamber is connected to the static system through a restricted (calibrated) leak. If the static pressure changes, the pressure in the diaphragm changes instantly, while the chamber pressure lags slightly behind. This causes the membrane to expand / retract which drives the needle on the instrument.

When the glass is broken, the chamber will be at the same pressure as the cockpit. Through the calibrated leak, the static system will be levelled with the cockpit pressure, but of course with a delay as well. The VSI will now work in opposite direction (if you didn't destroy the needle).

IVSI

The inner workings of an IVSI

Some VSI apparently have the calibrated leak into the cockpit already. For such VSI's smashing the glass does not work. Here the diaphragm should be destroyed as well.

Improve this answer
$\endgroup$
3
  • 6
    $\begingroup$ As the old adage goes, those in glass cockpits shouldn't thrown punches. $\endgroup$
    – corsiKa
    Oct 21, 2014 at 21:27
  • 1
    $\begingroup$ I wonder how many people would be able to work out if it was a dynamic or static problem they were facing? $\endgroup$
    – Simon
    Apr 4, 2015 at 21:18
  • 1
    $\begingroup$ @Simon: A blocked static port will cause the indicated altitude to freeze at what the altitude was when the port was blocked; a blocked pitot tube won't affect the altimeter. $\endgroup$
    – Vikki
    Jun 4, 2020 at 21:00
8
$\begingroup$

The FAA Instrument Pilot Handbook (11-8) mentions this as an alternate source if you lose your static source (not vacuum).

That said, I don't think every VSI is identical. I have seen arguments that some are not vented the same way and may not provide a valid static source.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Ach, thank you for clarifying that this is NOT for vacuum, but static. $\endgroup$
    – Bassinator
    Oct 20, 2014 at 23:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .