Understanding the Wedge Symbol in an Equation

In summary: It's not the same rotating frame as the aircraft body frame however. It's a frame that rotates about the aircraft z axis at the aircraft's yaw rate. This is why you need to be careful about frames. That ##\vec{\omega} \times (I \vec{\omega})## term is the Euler torque as seen in that gyro frame. This is how you make sense of the ##I \frac{d\vec{\omega}}{dt}## term. In that gyro frame, this is all the *visible* force working to change the aircraft's angular velocity.
  • #1
MHR-Love
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Hi guys,
Would you please help me by telling me the meaning of the wedge symbol in the equation attached?

I=inertia.
ω=angular velocity.

Since the inertia is constant, then why do we need to add the second term? I mean why isn't the differentiation of (Iω) simply equal to (Iω')?
 

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  • #2
Too long, didn't read answer: Abuse of notation, reference frames, and the transport theorem.

The left hand side of that equation would be better written as ##{\frac{d\vec L}{dt}}^{(I)}##, where the superscript (I) means "the time derivative of angular momentum from the perspective of an inertial frame."

Without proof, the transport theorem relates the time derivative of some vector quantity ##\vec q## as observed from the perspective of an inertial observer versus that of a rotating observer as
[tex]{\frac{d\vec q}{dt}}^{(I)} = {\frac{d\vec q}{dt}}^{(R)} + \vec\omega \times \vec q[/tex]

Use ##\vec q = \vec r## (i.e., position vector) and differentiate twice and you'll get the standard relation involving centrifugal and coriolis accelerations between accelerations as observed in a inertial frame versus that observed in a rotating frame.

What happens if you use ##\vec q = \vec L \equiv I\vec \omega\ ##? The left hand side, ##{\frac{d\vec L}{dt}}^{(I)}##, becomes the external torque. The right hand side becomes ##{\frac{d\vec L}{dt}}^{(R)} + \vec\omega\times \vec L##.

In any frame, ##\vec L = I\vec \omega##, but you have to beware that the inertia tensor I and the angular velocity ω are frame-dependent quantities. The inertia tensor of a rigid body is constant in a frame rotating with that body. Thus ##{\frac{d\vec L}{dt}}^{(R)} = {\frac{d}{dt}^{(R)} (I\vec \omega)}= I\frac{d\vec \omega}{dt}##. Note that I've dropped the superscript R from the final derivative because angular acceleration (time derivative of angular velocity) is the same vector in the inertial and rotating frame.

Putting this all together yields
[tex]
\vec \tau \equiv \frac{d}{dt}^{(I)}(I\vec \omega) =
I \frac {d\vec \omega}{dt} + \vec \omega\times(I\omega)
[/tex]
 
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  • #3
D H said:
Too long, didn't read answer: Abuse of notation, reference frames, and the transport theorem.

The left hand side of that equation would be better written as ##{\frac{d\vec L}{dt}}^{(I)}##, where the superscript (I) means "the time derivative of angular momentum from the perspective of an inertial frame."

Without proof, the transport theorem relates the time derivative of some vector quantity ##\vec q## as observed from the perspective of an inertial observer versus that of a rotating observer as
[tex]{\frac{d\vec q}{dt}}^{(I)} = {\frac{d\vec q}{dt}}^{(R)} + \vec\omega \times \vec q[/tex]

Use ##\vec q = \vec r## (i.e., position vector) and differentiate twice and you'll get the standard relation involving centrifugal and coriolis accelerations between accelerations as observed in a inertial frame versus that observed in a rotating frame.

What happens if you use ##\vec q = \vec L \equiv I\vec \omega\ ##? The left hand side, ##{\frac{d\vec L}{dt}}^{(I)}##, becomes the external torque. The right hand side becomes ##{\frac{d\vec L}{dt}}^{(R)} + \vec\omega\times \vec L##.

In any frame, ##\vec L = I\vec \omega##, but you have to beware that the inertia tensor I and the angular velocity ω are frame-dependent quantities. The inertia tensor of a rigid body is constant in a frame rotating with that body. Thus ##{\frac{d\vec L}{dt}}^{(R)} = {\frac{d}{dt}^{(R)} (I\vec \omega)}= I\frac{d\vec \omega}{dt}##. Note that I've dropped the superscript R from the final derivative because angular acceleration (time derivative of angular velocity) is the same vector in the inertial and rotating frame.

Putting this all together yields
[tex]
\vec \tau \equiv \frac{d}{dt}^{(I)}(I\vec \omega) =
I \frac {d\vec \omega}{dt} + \vec \omega\times(I\omega)
[/tex]

Thank you for the detailed reply. In reality, I am dealing with modelling an aircraft. ω is actually the body angular velocity of the craft and it is not the angular velocity of the rotating frame. I know that ## \vec \omega\times(I\omega) ## would represent the Coriolis effect and for finding it we need to know at what constant angular velocity the rotating frame is rotating. ω here is actually the craft angular velocity in its local frame not the velocity of the rotating frame. I read the following Wiki article and got some info. about the topic too. I might be confusing between frames and which ones I am to refer to.

http://en.wikipedia.org/wiki/Rotating_reference_frame
 
  • #4
You most certainly are working with a rotating frame. It's the body frame of your aircraft. The canonical aircraft body frame has its origin of the frame at the center of mass of the aircraft, and has axes such that the +x body axis points forward, the +y axis points the right wing, and the +z axis is x cross y to complete a right hand system.

There are two key issues with this frame. One is that the center of mass shifts as passengers and cargo are loaded or unloaded and as fuel is consumed. The other: How do you specify the location of the center of mass? Aircraft designers use a different but related frame to address these issues. The canonical aircraft structure frame has its origin a fixed distance in front of and below the nose of the aircraft. The structural +x axis points toward the rear rather than the front, parallel to the centerline of the aircraft. The structural +z axis points up rather than down, and +y is z cross x to complete a right hand system. The body and structural frames as described differ by a shift in origin and a 180 degree yaw.

You probably want the body frame rather than the structural frame because translational and rotational motion are decoupled in a center of mass frame. You do need to know of that other frame however because that's typically the frame in which center of mass is specified.

Regarding that ##\vec{\omega} \times (I \vec{\omega})## term, this is *not* the Coriolis effect. The Coriolis force is given by ##-2m\vec{\Omega} \times \vec v##. The Euler torque, as it is sometimes called, is given by ##-\vec{\omega} \times (I \vec{\omega})## (note the sign change). Here, ##I## is the aircraft's inertia tensor about the center of mass expressed in some body-fixed frame and ##\vec \omega## is the angular velocity of the aircraft with respect to inertial but expressed in the same frame as is the inertia tensor. That weird way of representing angular velocity is basically what your aircraft gyros report.

Your gyros actually report their output in a third body-fixed frame, the gyro case frame. The gyros don't know if you mounted them aligned with the body frame, upside down, or at some cockeyed angle. The gyro case frame x, y, and z axes are specific to the gyro. You know (or should know) how the gyro assembly is mounted to the aircraft, so you know (or should know) the relation between the gyro case frame and the aircraft structural and body frames. You can ignore this detail if you intentionally mount the gyros so that the case frame axes correspond to the (for example) body axes. It's still good to be aware of this the concept of a case frame. For example, you (or the aircraft designers) might have intended to mount the gyros so that they are perfectly aligned with the body axes, but nothing is perfect in reality. There's always going to be some misalignment. If you want to be very precise you should model this misalignment in your dynamics.
 
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  • #5


The wedge symbol in an equation typically represents the cross product between two vectors. In this case, the equation is representing the relationship between angular momentum (Iω) and torque (τ). The cross product between these two vectors is necessary because angular momentum and torque are both vector quantities, meaning they have both magnitude and direction. The cross product allows us to account for the direction of the torque and its effect on the angular momentum.

As for the second term in the equation, it represents the change in angular momentum over time. While inertia may be constant, angular velocity may not be. Therefore, the differentiation of (Iω) is not simply equal to (Iω'), as the change in angular velocity must also be taken into account. This is because the rate of change of angular momentum is affected by both the magnitude and direction of the torque, as well as the rate of change of angular velocity.

I hope this helps clarify the meaning of the wedge symbol and the necessity of the second term in the equation.
 

1. What does the wedge symbol represent in an equation?

The wedge symbol, also known as the caret symbol (^), represents the exponent or power in an equation. It is used to raise a number or variable to a certain power.

2. How is the wedge symbol different from the multiplication symbol?

While both symbols involve two numbers or variables, the wedge symbol represents exponential notation, while the multiplication symbol represents multiplication. The wedge symbol indicates that the number or variable before it is being raised to the power shown after it.

3. What are some common mistakes made when using the wedge symbol?

One common mistake is forgetting to use parentheses when raising multiple numbers or variables to a power. Another mistake is not correctly indicating the power, such as forgetting to include the exponent or using the wrong symbol.

4. How do I solve equations using the wedge symbol?

To solve an equation with the wedge symbol, you can use the laws of exponents. For example, when raising a power to another power, you can multiply the exponents. It is also important to follow the order of operations when solving equations with multiple operations and the wedge symbol.

5. Can the wedge symbol be used with fractions or negative numbers?

Yes, the wedge symbol can be used with fractions and negative numbers. When working with fractions, both the numerator and denominator can be raised to the power indicated by the wedge symbol. When working with negative numbers, the entire number, including the negative sign, is raised to the power shown after the wedge symbol.

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