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I am reading this slide: enter image description here

Source: http://www.stengel.mycpanel.princeton.edu/MAE331Lecture10.pdf

Is Izx always considered small and we can assume it is 0, as many texts assume?

I need to make this clear: This is the body axis: enter image description here

I assume that G is the center of mass, GZX is the symmetry plane. So Iyz = Iyx = 0. But why is Ixz zero? Do they rotate the body coordinate system around Y axis so that Ixz = 0?

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  • $\begingroup$ I think you may get a quick and straight answer by emailing the lecturer (just a suggestion, nothing against your question) $\endgroup$ – Afe May 6 '19 at 11:49
  • $\begingroup$ It does appear to be a standard term (I find it in several other publications), and I can even approach the calculation. But nothing I found gave a more intuitive interpretation of the element. $\endgroup$ – BowlOfRed May 7 '19 at 20:12
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This is an addendum to @ROIMaison's answer based on the modified OP.

I assume that G is the center of mass, GZX is the symmetry plane. So Iyz = Iyx = 0. But why is Ixz zero? Do they rotate the body coordinate system around Y axis so that Ixz = 0?

The coordinate axes are completely arbitrary. By convention, the x-axis aligns on the symmetry plane and points from nose to tail. However, one can choose literally any orientation. You can, of course, align the axes such that all the cross inertial products are zero; such axes would be the principle axes. But obviously this would only work for a single configuration, fuel distribution, etc.

To answer your question more succinctly, no, we don't change the axis once the initial drawings are made. The coordinate system stays fixed and becomes the body axis. For a conventional design, the cross products would be fairly small and would not be a big deal to ignore for first order flight dynamics analysis.

What would happen if you align the axes weirdly such that the cross products are all large and non-zero? Nothing really, it just complicates your math and modeling, and your angular velocity $[p,q,r]$ would be hella confusing. But physics would work out just right.

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  • $\begingroup$ so your answer is that: Ixz is small so we can set it to be zero. What about the author idea of nose high/low product of inertia ? What does that mean? $\endgroup$ – Dat Oct 5 '19 at 6:43
  • $\begingroup$ Nose tilted higher (nose-high) than the x-axis, or nose tilted lower (nose-low) than the x-axis, produces cross inertial product. Simple as that. $\endgroup$ – JZYL Oct 5 '19 at 6:45
  • $\begingroup$ that is so obvious, that confuses me why the author said such obvious thing $\endgroup$ – Dat Oct 5 '19 at 7:05
  • $\begingroup$ It's a name given by the author for Ixz. $\endgroup$ – JZYL Oct 5 '19 at 7:18
  • $\begingroup$ wow that makes sense to me now $\endgroup$ – Dat Oct 5 '19 at 9:11
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The $I_{xz}$ moment of inertia is a measure for how the weight is distributed in the xz plane.

You can obtain it by multiplying a unit of mass $dm$ by its distance in $x$ and $z$ from the center of gravity and taking the negative of it.

For discrete masses (blocks of mass) it looks like this: $$ I_{xz} = \{-(x_1 \cdot z_1) \cdot m_1\} + \{-(x_2 \cdot z_2) \cdot m_2 \} + \{-(x_3 \cdot z_3) \cdot m_3 \} ... $$

If you look at the plane in question, you can see that it has a nose - which we can consider as one of the blocks - which is

  • Far in front of the c.o.g. (large value of $x$)
  • High up from the c.o.g. (large value of $z$)

If we plug this in the formula, we see it leads to large values of $I_{xz}$. Similarly, there is also a lot of mass in the lower right corner

  • Far behind the c.o.g. (large negative value of $x$)
  • Low below the c.o.g. (large negative value of $z$)

In essence, everything that is on the x-z line contributes to this value, and as we see from the marked area in the image of the Valkyrie, there is a lot of mass around this line.

If your aircraft is symmetrical around the x or z axis (in terms of as much mass above/below front/aft the axis) you will get a low value of $I_{xz}$.

I guess that is the case for many aircraft, so they make the assumption of $I_{xz}=0$.

More information about the rotational moment of inertia can be found here.

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    $\begingroup$ To be more precise, if the aircraft is symmetrical about x OR z axis, the cross product will be zero. $\endgroup$ – JZYL Sep 30 '19 at 13:22
  • $\begingroup$ Ah good catch, I will change it, thanks! $\endgroup$ – ROIMaison Sep 30 '19 at 13:30
  • $\begingroup$ Would the anonymous downvoter care to elaborate? $\endgroup$ – ROIMaison Oct 1 '19 at 7:50
  • $\begingroup$ if we have a block that has large value of x and large value of z, and a block that has large negative value of x and large negative value of z then the total of them would be larger, they do not cancel out each other like the idea in the answer. Please explain this point further $\endgroup$ – Dat Oct 2 '19 at 7:41
  • $\begingroup$ @Dat In the answer, I mention that geometries that are symmetrical around $x$ or $z$ have $I_{xz}=0$. The case in your comment is neither $\endgroup$ – ROIMaison Oct 2 '19 at 8:57
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Note that the inertias are defined in the body axis system. Not all the main inertial axes are aligned with the body axis system: The high nose section and the low engines combine to tilt the inertial X axis nose-high. This requires a correction term in the inertial matrix, namely the I$_{xz}$ term.

If, say, the left wing would be heavier than the right wing and the right canard much heavier than the left canard, there would be an additional, nonzero I$_{xy}$ term. Due to the left-right symmetry this term is also zero for the XB-70. More generally, if the aircraft mass distribution can be represented by the sketch below, the cross-coupling terms (those with unequal indices) are nonzero.

unequal mass distribution

Unequal mass distribution: The long rod represents the fuselage with its X-axis (dot and dash line) while the cg symbol represents an additional mass mounted at a distance from the X-axis. Without that additional mass, or if the mass would be at the center of the rod, the I$_{xz}$ term would be zero indeed. You need both: Distance in X and in Z for it to add I$_{xz}$.

Now imagine the arrangement above is rotating around the X-axis: The centrifugal force on the additional mass will add a strong pitching moment, pulling the arrangement into a vertical orientation. This does not happen without that extra mass or without a lever arm of that mass in X direction.

Most aircraft don't have this misalignment between body and inertial X-axes, so for them it is correct to set I$_{xz}$ to zero. Notable exceptions are the Douglas X-3 and a range of McDonnell-Douglas aircraft like the F-101 Voodoo and the F-4 Phantom II: Here the inertial x-axis was pointing down relative to the body axis and artificial stability augmentation had to be added to avoid roll inertia coupling.

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  • $\begingroup$ Ixz is generally small compared to the other principle inertial terms, and are ignored for first order studies. But it is never actually zero unless you have a flying saucer. $\endgroup$ – JZYL Oct 1 '19 at 5:26
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    $\begingroup$ @Jimmy: Aviation is engineering, so close enough to zero is zero. Also please note that I never said that I$_{xz}$ is zero, that is only your interpretation. $\endgroup$ – Peter Kämpf Oct 1 '19 at 10:02
  • $\begingroup$ @PeterKämpf: I always assume all aircrafts are symmetrical around xz plane so for me, Ixy is always zero (no doubt). Why did you metion Ixy here? I don't get your point. All I want to know is why Izx = 0 ? $\endgroup$ – Dat Oct 3 '19 at 2:23
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    $\begingroup$ @Dat: I added an example in a different plane, that is all. Sorry if this confuses the matter more than it helps. I added more on the xz asymmetry - let me know if this is more helpful or not. $\endgroup$ – Peter Kämpf Oct 3 '19 at 8:08
  • $\begingroup$ @PeterKämpf I added some information in the post above about where I am stuck $\endgroup$ – Dat Oct 5 '19 at 2:26

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