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I am interested in improving the in-flight behavior of an oblong projectile that is launched from something like a trebuchet (a rotating arm).* This is a hobby project and I'm not an engineer or physicist.

My problem is that the oblong projectile tends to tumble. I want it to fly straight like an arrow.**

What I want to know is simply where the center of gravity of an oblong projectile should be given my parameters.

Those parameters are:

  • the launch has a rotational component that will tend to make the oblong projectile tumble
  • speeds are relatively slow: max of 150 feet per second, and usually half that.
  • distances are relatively short: max of 100 feet, and typically half or a quarter of that.
  • the oblong projectile is very dense: it's solid steel and weighs about 250 grams, has length of 250 mm and diameter on the order of 10 mm (approx).
  • the oblong projectile has very minimal spin around its longitudinal (leading to trailing) axis, and it is not possible to give it more spin (such as a rifled bullet has).
  • the oblong projectile has no fins or stabilizer -- it's cylinder shaped with a point at the front. Think of a small torpedo, missile or spear without any fins. The cylinder does not have to be constant diameter nor have straight sides.
  • air resistance is negligible (not worth considering); assume, fins would not help even if they could be added (which they can't).

I mention specific speeds, lengths, etc. only to give an example. The problem is a general one. The important conclusion from the above parameters is that solving this problem cannot rely on the usual methods such as center of pressure being behind the center of gravity. A different approach is required.

My question is simply: should the center of gravity to be toward the front, balanced or toward the rear? (And, if possible, why?)

UPDATE: The projectile is positioned radially on the launcher with the "front" furthest from the pivot of rotation and the back closest to the pivot. Therefore, the front of the projectile has higher velocity (compared to the rear) during the launch. It's similar to a spoke in a wheel, where in this case, the front tip of the projectile is mounted near the rim with the back end toward the hub. When the projectile is released from the rotating arm, it should fly straight without tumbling. I will work separately on engineering issues related to the release of the projectile from the rotating arm. For this general problem, I only wish to know where the center of gravity should be given that air/fluid pressure won't help stabilize the projectile.

*NOTE 1: I have tried to simplify the description to avoid a long distracting explanation of how the projectile is thrown. I'm not using a trebuchet. I'm using something with more pivots. But I want to keep the question general and simple. Engineering details of the launcher are outside the scope of this community discussion.

**NOTE 2: a lot of the tumbling will relate to the characteristics of the launch (the throw). But, again, those issues are outside the scope of this question.

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  • $\begingroup$ This seems to be an engineering question - i.e. about the solution of a specific real world problem using physics - rather than a question about general physical principles. Would it be better on Engineering? $\endgroup$ – ACuriousMind Jul 27 '17 at 9:33
  • $\begingroup$ @ACuriousMind: if that's what StackExchange Physics suggests, I'll try it on Engineering. A person who has worked on this told me, "it's a physics problem" which is why I started here. :-) $\endgroup$ – MountainX Jul 27 '17 at 9:35
  • $\begingroup$ I see no close votes on this yet - and I think it's debatable enough to not close it unilaterally myself. $\endgroup$ – ACuriousMind Jul 27 '17 at 9:47
  • $\begingroup$ I asked again and I'm confident this is a general problem, not a specific engineering problem. The answer will be the same regardless of the specific length, diameter, mass or material of the projectile. I gave specifics because I thought people might want those. But I could leave out all specifics and the question would be just as valid to me. It's a general question that stands apart from any specific implementation / real world solution. And, I think, it is a very challenging problem that is a lot more complex than it appears at first. So, thanks for not closing it. :-) $\endgroup$ – MountainX Jul 28 '17 at 3:22
  • $\begingroup$ Possibly related : Why do throwing knives need to be balanced? and Stability of rotation of a rectangular prism $\endgroup$ – sammy gerbil Jul 29 '17 at 16:43
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If you want your projectile to rotate less, the easiest way to do so (in the absence of air resistance) is to make it difficult to rotate. This is equivalent to maximizing the moment of inertia of your projectile when rotating on an arbitrary axis through the center of mass and perpendicular to the travel direction. Since your projectile has cylindrical symmetry, it doesn't matter which particular axis we pick, as long as it's perpendicular to the axis of the cylinder.

There are generally three different ways we could proceed:

  1. Taper heavily toward the back,

  2. Taper heavily toward the front, or

  3. Taper in both directions, creating a dumbbell shape.

Let's examine the limit of these three cases. In case 1 and 2, we get a thin disk* (i.e. basically all of the mass is at the front or back), and in case 3 we get two thin disks (one at each end). Assume each case has the same total mass $M$ and total area $A$.

Cases 1 and 2 (taper to front or back)

Cases 1 and 2 are single disks with radius $R=\sqrt{\frac{A}{\pi}}$, so their moment of inertia is

$$I_1=I_2=\frac{MR^2}{4}=\frac{MA}{4\pi}$$

Case 3 (make a dumbbell)

The radius of the disks in case 3 are such that $2\pi R^2=A$, so $R=\sqrt{\frac{A}{2\pi}}$ in this case. The moment of inertia of each disk is $\frac{MR^2}{4}=\frac{MA}{8\pi}$ when rotated about its center of mass. But now we apply the parallel axis theorem, which moves the disks' axis of rotation a distance $L/2$ away (where $L$ is the total length of the projectile). So the moment of inertia of each disk about this new axis is $\frac{MA}{8\pi}+M\left(\frac{L}{2}\right)^2$ by the parallel axis theorem, and there are 2 identical disks, so the total moment of inertia is

$$I_3=2\left(\frac{MA}{8\pi}+\frac{ML^2}{4}\right)=\frac{MA}{4\pi}+\frac{ML^2}{2}$$

Thus, we see that in every case, the moment of inertia in case 3 is greater than that in case 1 or 2. So, in the absence of air resistance, making the projectile dumbbell-shaped is the best option.

*It may be counterintuitive that tapering toward the front and the back would be the same, but note that a front-tapered projectile is just a back-tapered one rotated by 180 degrees, so they would be the same object.

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  • $\begingroup$ How would the answer differ considering that the launcher already starts the projectile rotating and we wish to have that rotation stop during the flight? In my understanding, maximizing the moment of inertia will cause the already rotating projectile to keep rotating. I want the opposite of that. $\endgroup$ – MountainX Jul 27 '17 at 9:19
  • $\begingroup$ @MountainX In the absence of air resistance, there are no torques on the object in-flight. Therefore, there is no way to stop any rotation that is imbued by the launch. A higher moment of inertia means that the rotation imbued by the launch is slower; this is the best we can do without air resistance to help us. $\endgroup$ – probably_someone Jul 27 '17 at 9:23
  • $\begingroup$ Second point. The launcher is a rotational arm and the projectile is positioned with the "front" furthest from the pivot of rotation and the back closest to the pivot. Therefore, the front of the projectile has higher angular velocity during the launch. Considering this, it does make a difference how we orient the projectile on the launcher. My original question can thus be stated: if we have a heavily tapered projectile, should we mount the heavy end furthest from the pivot of rotation or closest to that pivot? $\endgroup$ – MountainX Jul 27 '17 at 9:26
  • $\begingroup$ Here's my theory: given the rotational nature of the launch and the front of the projectile having higher angular velocity, the back should be heavier. This should reduce the tendency to rotate... I guess. $\endgroup$ – MountainX Jul 27 '17 at 9:29
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I would like to know more about the geometry of your setup, but failing that, here is some elementary aerodynamics of things shaped like missiles. That is, a pointed nose faired into a cylinder, with the thing travelling roughly parallel to the axis of the cylinder. According to well-established fluid flow theories, in inviscid flow there is no transverse force acting on the cylindrical portion of the body. There is a transverse force on the nose. Therefore there is a strong moment about any point aft of the nose. For stable flight, the c.g. must be ahead of the "center of pressure", and hence in the front portion of the nose. This is why all missiles have stabilising fins at the rear, just as arrows do. Moments of inertia have to do with how rapidly the unstable missile will rotate, but not much to do with whether it will rotate. Viscous effects will tend to add stability but not to a useful extent. You say that you can neglect air resistance, but you cannot neglect transverse aerodynamic forces because they have no competition.

Presumably you have thought about stabiliser fins and concluded that they are not easily accommodated into your launch system. In that case you have no hope without fancy stuff such as gyroscopes. Can you devise fins that fold away during launch but that somehow spring out afterwards?

To fly like an arrow, it will need to look like an arrow!

EDIT I just recalled that you said the projectile would be very dense, and tumbling when launched. This is making things very difficult. The aerodynamic forces are the only ones available to straighten up the flight, and must be comparable to the inertia that you want to remove. In other words the fins must be big.

It is not impossible to have them do the job. Consider the game of darts. You can toss a dart at random and it will straighten out. But take scissors and trim the flights. They need to be big enough.

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  • $\begingroup$ Here's my reasoning. The front of the projectile has greater angular velocity. If the cg is toward the rear, the rear will absorb / resist the rotational energy imparted to the front. Exploiting the combination / interaction of differing mass at each end and differing angular velocity at each end is the only way I can think of to reduce the tendency to rotate (given that I can't use fins, gyroscopes, etc.). Anything to this? (It seems to possibly be working in some early tests...) $\endgroup$ – MountainX Jul 27 '17 at 21:31
  • $\begingroup$ If we are talking about a rigid body, every part of it has the same angular velocity, doesnt it? $\endgroup$ – Philip Roe Jul 27 '17 at 22:45
  • $\begingroup$ Maybe I have my terminology wrong. Maybe I should have said velocity, not angular velocity. Sorry. For a bicycle wheel, the tire has greater velocity than the hub. In this case the front tip of the projectile has more velocity than the rear. It is mounted radially like a spoke in a wheel. $\endgroup$ – MountainX Jul 28 '17 at 1:03
  • $\begingroup$ So when it is launched, it will be spinning about its c.g. and in the absence of any applied forces, it will continue so. The only forces available are going to be aerodynamic. $\endgroup$ – Philip Roe Jul 28 '17 at 3:57

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