I am trying to design a VTOL system that uses a servo to rotate a propeller and the attached motor so that it can provide thrust when it is at 0 degree and lift when it is at 90 degree. However, I understand that when the propeller is rotating, it behaves like a gyroscope and if the propeller starts to rotate, there is got to be stresses developed at both the propeller blades and the propeller shaft connections. But I have no idea how to start forming an equation.
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1$\begingroup$ I think CAD softwares provide a value of moment of inertia (about all the axes) if you the geometry available. $\endgroup$– Rameez Ul HaqJan 21, 2022 at 20:47
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$\begingroup$ Note that for a two-blade propeller, vibrations are induced when tilting the propeller disc because the moment of inertia about the tilt axis changes throughout the spinning of the propeller between a maximum and a minimum since the propeller is not radially symmetrical. Three-blades is the minimum to not have this happen (or greatly reduce it). supercoolprops.com/articles/gyrovibes.php $\endgroup$– DKNguyenJan 21, 2022 at 20:53
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3 Answers
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In order to calculate the mass moment of inertia accurately your best bet is using the functions of a CAD system (solidworks, inventor, onshape etc).
If all you want is an an approximation and you can afford the assumption that each blade of the propeller is a long bar with made from a single material (or that the density is uniform along the length of the blade), then you could calculate the mass moment of inertia about the rotation axis as a sum of all3 blades and the hub.
blade
With the above assumption (i.e. long bar of uniform density) then the mass moment of rotation about the rotation axis is:
$$I_{b} = \frac{1}{3} m_b \cdot r_b^2$$
where
- $r_b$ is the length of the blade
- $m_b$ is the mass of the blade
**Figure: blade approximation as a rotating bar (source: wikipedia) **
hub
the blade can be approximated as a rotating disk of mass $m_h$, and radius $r_h$, and in that case the mass moment of inertia would be:
$$I_h =\frac{1}{2} m_h \cdot r_h^2$$
Total
The total mass could be approximated by:
$$I_h + n_b\cdot I_{b} = \frac{1}{2} m_h \cdot r_h^2 + n_b\cdot \frac{1}{3} m_b \cdot r_b^2 $$
where: $n_b$ is the number of blades on the propeller.
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$\begingroup$ But this is only when the prop is stationary, right, what about when it is running? $\endgroup$ Jan 22, 2022 at 19:22
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$\begingroup$ The mass moment of area will not change (at least not about the axis of rotation). $\endgroup$– NMech ♦Jan 22, 2022 at 19:27
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$\begingroup$ So it has the same moment of inertia when it is stationary and when it is moving? But it would behave like a gyroscope when the prop is rotating, right? Or else why gyroscope can produce resistance to turning? $\endgroup$ Jan 22, 2022 at 19:37
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$\begingroup$ The mass moment of area is similar to a geometric characteristic like the area of a cross-section. If you rotate it about an axis it doensn't change. The resistance of the gyroscope is due to the preservation of the angular momentum which is another phenomenon altogether. $\endgroup$– NMech ♦Jan 22, 2022 at 19:39
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$\begingroup$ No, the entire setup--motor and prop--rotates from facing the front to facing the top $\endgroup$ Jan 23, 2022 at 2:26
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Assuming you already know basic moment of inertia calculations (and if you do not then you really should not be asking your question here prior to research):
You can't develop an equation without knowing the propeller geometry, at which point you would not be able to do it by hand. If you want to do it by hand then chop the propeller blades up into slices and weigh each slice and record it's distance from the center. Then you can model it as a bunch of point masses along a line using regular old moment of inertia principles.
Tables are provided to calculate the moment of inertia of geometric objects. These also let you calculate moment of inertia by modelling real objects as a composite of geometric objects. But propellers, being inconsistently tapered objects as they are, are not geometric objects. Though you could model the hub as a cylinder or a ring, you probably do not want to approximate propeller blades as a beam or square/rectangular prism.
In any case, one of the things you will need to know the parallel-axis thereom which is part of the basics of moment of inertia calculations.
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When the propeller is rotating, it creates a torque which results in shearing stress on the connection. What you need is the "mass moment of inertia" (rotational inertia), $I = mr^2$. In which, $m$ is the mass (lumped masses) of the propeller, "$r$" is the radius of rotation about the shaft.
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3$\begingroup$ That equation assumes the entire mass of the propeller is at its tip since it is really for point masses rotating around a radius. $\endgroup$– DKNguyenJan 21, 2022 at 21:08
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$\begingroup$ @DKNguyen Please review the concept of inertia force of an object with arbitrary shape in motion - where the force is applied, and how to measure the distance (arm) between the force and the reference datum. $\endgroup$– r13Jan 21, 2022 at 21:47
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3$\begingroup$ You're going to have to be more explicit about what you mean. There's no moments of inertia in the linked post or any required distinctions between center of masses such as that of the whole propeller versus just the blades. $\endgroup$– DKNguyenJan 21, 2022 at 23:21
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3$\begingroup$ Oh, are you referring to the non-mathematical equal sign? I would almost say it doesn't need to be as precise as the textboook but it shouldn't be even more misleading by omission textbooks often are. Not mentioning what radius just leaves too much unsaid, especially when it is the crux of this problem. And this crux is further obfuscated by saying something like "lumped mass of the propeller" which strongly draws one away of the COM of the blades towards the COM of the propeller (i.e. the hub). $\endgroup$– DKNguyenJan 22, 2022 at 0:07
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1$\begingroup$ When I wrote that answer, I actually was pondering whether I should have used a colon but chose the equal sign because if you look at it, many of the the other definitions it leads into equations. A colon preceding text descriptions of the variable would have broken that chain. $\endgroup$– DKNguyenJan 22, 2022 at 0:11