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Gyro's - Turn Indicator

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  • Gyro's - Turn Indicator

    After just completing Progress Test 2 on the Instruments subject I have a small query regarding one of the questions I got wrong.

    The question asked: If a turn slip indicator overspeeds the result will be:

    I gave the answer as the indicator will under read

    The correct answer given was that it will over indicate the turn.

    I understand your reasoning, that because the gyro underreads when it is stopped, then while it is winding down it will underread. Therefore if it overspeeds it will over read. This I believe is possibly incorrect.

    A spinning gyro at rest is not a gyro. It does not hold the attributes of a gyro. Therefore the rules applied to a gyro can not be applied to a gyro that is stationary.

    Newtons second law states in Formula the following:

    Force = Momentum x Acceleration.

    This is essentially the property of a rigid gyro. Surely from this formula we can deduce that if, through the angular momentum of a gyro, the speed increases (acceleration), then in order to keep the formula equal, the momentum increase thus the force required to change the direction of the gyro must be increased. If this is true, then the speeding up of a gyro means that a greater force is required to change it's direction. There fore if a aircraft is in a rate one turn, which is applying a certain force, if that gyro now decides to speed up in that turn then with the increase of momentum and increase in accelaration of the gyro unless there is a greater force to oppose this increase it will naturally want to errect itself to its original position i.e. underread. To put it more simply, it is a lot more diffcult to turn your bicycle wheel when it is turning at 2000rpm as opposed to turning at 100 rpm. Therefore in a turn the indicator will underread as the force or turn required would have to be greater to get the indication required.

    Maybe i am wrong.........Please clarify.


    Regards
    iroquois

  • #2
    Hi,

    The turn indicator is calibrated on the assumption that the gyro will be rotating at its design rpm. If it is rotating at its design rpm and you turn at 3?/sec it will indicate a rate one turn. If the gyro is rotating at a slower rpm and you turn at rate 1 there will be less precession of the gyro and it will under-indicate the rate of turn. Similarly, if the gyro is rotating at more than its design rpm there will be more precession for the same yaw force and the instrument will over-read the rate of turn. The amount of precession is proportional to the angular momentum (rpm x mass) and must not be confused with the property of rigidity which simply states that with high angular momentum the gyro is likely to stay where it is if it is left alone.

    Regards
    Baz Hamblin

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    • #3
      Of course yes.......

      I was looking at it from a totally different perspective.

      Thank you for your kind explanation

      Regards

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      • #4
        Hi. i am confused by this. If the gyro is rotating at a slower rpm and you turn at rate 1 there will be less precession of the gyro
        I thought the precession would be inversely proportional to the rpm. so a slower rpm would imply more precession. is that not true? thanks

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