Why do bullets precess in the opposite way from gyroscope diagrams?

[johnlpmark]

TL;DR Summary

Rifled bullets are gyroscopes: precession maintains the center of pressure before the center of mass. A side effect is deflection from yawing (aerodynamic jump) and pitching (spin drift). Spinning bullets and gyroscope diagrams show opposite forces.

1) Bullet spin causes a bullet to become a gyroscope. Specifically, bullets have their center of pressure in front of their center of mass. Therefore, when pressed, gyroscopic forces cause a bullet to spin 90 degrees instead of tumble. See this diagram:

2)The precession does not cause the bullet to point into the direction of movement, but the direction of apparent incoming wind. That means that if there is a crosswind of 20mph, the bullet will turn slightly to point towards the incoming wind while traveling straight forward. See this diagram:

3a) As a bullet drops, there is apparent wind coming from the downward direction. Apparent wind occurs when there is a differential between the object and the surrounding medium. Therefore, the bullet turns down as it drops. See this diagram below.

3b) Importantly, gravity accelerates, so there is always an increasing apparent wind as the bullet falls. This accelerating vertical apparent wind causes continuous forward rotation of the bullet.

4a) Air resistance forces to the front of the bullet and those to the side of the bullet cause different behavior. Air resistance to the front causes the bullet to reorient into the direction of the incoming force. (See the first image.)

4b)Air resistance to the side of the bullet (i.e., from the left, up, down, or right directions) pushes the bullet to point 90 degrees clockwise to the incoming air. So left-to-right crosswind cause aerodynamic jump down and a or right-to-left crosswind causes aerodynamic jump up. (Note the horizontal defection in this image is due to the crosswind, which is distinct from spin drift.) See this diagram:

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5) Therefore, intuitively, when a bullet rotates down due to being in a falling trajectory with gravity, it points to the right if it has a right-hand twist:

6) But every diagram I see of a gyroscope has the gyroscope rotating the opposite direction.

7) I can’t figure out why the bullet gyroscope rotates in the opposite direction of other gyroscopes. This is not Magnus effect! From Modern Exterior Ballistics, McCoy, “Although the Magnus force acting on a spinning projectile is usually small enough to be neglected, the Magnus moment must always be considered.” Note in the diagram that the Magnus moment points down, and not the to left or right. See these diagrams:

8) In contrast, spin drift is related to the pitching moment, which does point left or right. See the below diagram. Importantly, bullets do not roll off of air. Indeed, if that were the case, then the rear has more surface area and it would cause the bullet to point away from the incoming air flow which is not the case.

9) Something to note is that the traditional depiction of bullet rotation may be somewhat inaccurate. (I cannot confirm.) Traditionally bullets are depicted as rotating around the center or mass. Instead, they may rotate around an independent axis. See this diagram:

Thanks so much for the help!

-John

 

[A.T.]

5) Therefore, intuitively, when a bullet rotates down due to being in a falling trajectory with gravity, it points to the right if it has a right-hand twist:

The aerodynamic torque in that diagram has the wrong direction.

 

[johnlpmark]

I suppose the label "aerodynamic" is misleading, as you are correct, the lift force causes torque in the opposite direction

However, it is spin-stabilization/gyroscopic-precession that redirects the lift force into an aligning force. Without the green arrow, the entire concept of spin-stabilization falls apart, as spin-stabilization by definition orients the bullet into the apparent incoming direction of airflow.

Perhaps "net vertical torque" would be a better label?

 

[johnlpmark]

The bullet tips into the direction of the incoming air (called weather-vaning). From the perspective of the bullet, this is a forward rotation.
...
.... a right spin should force a forward rotation to induce a left turn.

You seem to conflate the change of the bullets orientation with the aerodynamic torque acting on the bullet. You call both "forward rotation", but in the second instance you seem to actually mean the aerodynamic torque, and its gyroscopic interaction with the spin.

For fin-stabilized-projectiles this aerodynamic torque would indeed act to push the nose down (forward rotation). But for spin-stabilized projectiles this is not necessarily the case, as the diagram you posted shows. Here the drag force R passes above the center of mass C, thus acting to lift the nose (backward rotation):

You also posted this on my previous inquiry. Thank you for responding! First, I'm not sure that "R" refers to drag; usually that is Fd (force of drag). I think the arrow is supposed to point sideways, not backwards.

Also, why would a component forces (lift force) be more relevant than net force? Isn't the final change in orientation indicative of the total net force or torque? In the image, "C" is what occurs: bullets orient into the apparent path of incoming air and hence must experience corresponding forward rotation/torque. Much appreciated.

 

[Baluncore]

Without the green arrow, the entire concept of spin-stabilization falls apart, as spin-stabilization by definition orients the bullet into the apparent incoming direction of airflow.

Where is that definition?

What you are suggesting would require the bullet have tail fins, or fletches, and that it not spin, so it can fly parallel with the trajectory, like an arrow.

Spin stabilization causes the bullet to maintain its axial orientation, while the airflow around the bullet is not aligned exactly with the bullet axis. That causes the bullet to walk sideways in flight.

 

[johnlpmark]

"Spin stabilization causes the bullet to maintain its axial orientation." This is not true in any reputable source. What references are acceptable to you so I can provide them?

 

[johnlpmark]

Static stability does not maintain the gyroscope in its orientation in space, but maintains the orientation of the gyroscope in the direction of of the overturning moment. - http://ffden-2.phys.uaf.edu/212fall2001_Web_projects/Isaac Rowland/Ballistics/Bulletflight/stab.htm

I think McCoy, Modern Exterior Ballistics is also a good reference, although very difficult to parse.

Page 231 "Section title: 10.6 The yaw of repose for spin-stabilized projectiles - The yaw of repose for spin-stabilized projectiles is given the particular solution of equation (10.79), and it is caused by the interaction of axial spin and the acceleration due to gravity. Gravity pulls the projectile downwards, and the projectile's spin axis always lags very slightly behind. The result is a tiny overturning torque, caused by the pitching moment, that tries to rotate the shell's nose upward. But the spin-axis of a gyroscope rotates in the direction of the vector applied torque (Ref. 1)."

Perhaps the misunderstanding is in the word "maintain." Gyroscopes resist turning. But they do turn (albeit slower than a non-stabilized item) and do not maintain their original spin-axis. The spin-axis does reorient.

 

[Baluncore]

Without the green arrow, the entire concept of spin-stabilization falls apart, as spin-stabilization by definition orients the bullet into the apparent incoming direction of airflow.

I disputed that statement and asked you for the definition of spin stabilisation that you have assumed.

"Spin stabilization causes the bullet to maintain its axial orientation." This is not true in any reputable source. What references are acceptable to you so I can provide them?

The primary reason for spin stabilisation is to maintain a stable axial orientation.
There are secondary effects of the spin that come into play during flight.

You are making this discussion too complex.
I will avoid responding to your anger and emotion.

 

[johnlpmark]

The purpose of spin stabilization is to maintain the orientation of the bullet in a way that reduces the drag force.

I was referencing the fact that if the bullet maintained its orientation in space for the entire trajectory, then the orientation of the bullet would actually be detrimental at long distances. This is because facing perpendicular to the incoming airflow would create a larger surface area than simple tumbling or a spherical round.

This is less evident in rifle rounds, but is very evident in artillery rounds which are fired at much greater angles such as 45 degrees. Check out these undisturbed, unexploded WW1 and WW2 artillery rounds, which were all ended up with some degree of downward facing and did not have fins.
https://www.alamy.com/an-unexploded...?imageid=B6130860-9687-4CD7-8DA1-79EF2722E362
https://www.alamy.com/unexploded-ge...?imageid=F968A924-14A4-47AE-8DD6-C95B8857E1B0
https://www.alamy.com/stock-photo-a...?imageid=D8D3E8D0-E8FC-449D-95F7-3BF1E29B4577

In terms of "falling apart," this is where the complexity comes in. There are two kinds of stabilization in bullets, static and dynamic. Static stabilization is gyroscopic stabilization. Dynamic stabilization is the ability of the bullet to decrease the angle of yaw.

McCoy page 233 "If either damping exponent becomes (and remains) positive, that yaw mode will begin to grow, which is the definition of instability."

The angle of yaw is relative to the trajectory, therefore, by McCoy's definition, a dynamically stable bullet aligns to the orientation of the spin axis. And the spin axis changes due to outside torque.

I'm assuming this refers to the magnus force "That causes the bullet to walk sideways in flight."? If not please clarify. The magnus force is does not significantly contribute to the sideways movement of bullets, and I will city McCoy “Although the Magnus force acting on a spinning projectile is usually small enough to be neglected, the Magnus moment must always be considered.” Can you please expand on what this means? Thank you.

 

[Baluncore]

The purpose of spin stabilization is to maintain the orientation of the bullet in a way that reduces the drag force.

That is not a definition of spin stabilization, it is a purpose.

The complexity arises because you cannot be concise.

 

[johnlpmark]

If 301 words is not concise enough, or you do not like the complexity then we are at an impasse. I have cited McCoy as you requested. In turn, I am requesting that you cite or clarify this statement "That causes the bullet to walk sideways in flight." Thank you.

 

[A.T.]

I suppose the label "aerodynamic" is misleading, as you are correct, the lift force causes torque in the opposite direction

Not just lift. Even pure drag force at that CoP would pass above the CoM and act to lift the nose.

However, it is spin-stabilization/gyroscopic-precession that redirects the lift force into an aligning force. Without the green arrow, the entire concept of spin-stabilization falls apart, as spin-stabilization by definition orients the bullet into the apparent incoming direction of airflow.

Gyroscopic-precession will cause the spin axis to precess around the apparent incoming direction of airflow, not create a torque towards it.

Perhaps "net vertical torque" would be a better label?

No, there is no such torque.

 

[A.T.]

For fin-stabilized-projectiles this aerodynamic torque would indeed act to push the nose down (forward rotation). But for spin-stabilized projectiles this is not necessarily the case, as the diagram you posted shows. Here the drag force R passes above the center of mass C, thus acting to lift the nose (backward rotation):

First, I'm not sure that "R" refers to drag; usually that is Fd (force of drag).

You originally posted that picture. You should provide the definition of the labels. I assumed R = resistance

I think the arrow is supposed to point sideways, not backwards.

If R points towards the viewer then it could the aerodynamic torque acting to lift the nose.

Also, why would a component forces (lift force) be more relevant than net force?

The total aerodynamic force (lift and drag) is what creates the aerodynamic torque around the CoM. Gravity doesn't create a torque around the CoM.

bullets orient into the apparent path of incoming air and hence must experience corresponding forward rotation/torque.

No there is no such "forward torque". The spin axis simply precesses according to the aerodynamic torque. There is no "new torque" created, that rotates the spin axis. You seem to have some misconceptions about gyroscopic-precession.

 

[johnlpmark]

Haha, I absolutely do not fully understand precession. I appreciate the responses. How does the spin axis reorient in a forward rotation to the incoming airflow without it being called a torque? I think this is key and I do not understand.

While gravity does not apply a torque directly, it is the indirect cause of changing lift and drag forces over time. I should have been clearer. The reason I pointed this out is to say that since gravity is an accelerating force (quasi-force), so too the aerodynamic torque also accelerates.

My bad for the diagram, I should have provided labels.

"Not just lift. Even pure drag force at that CoP would pass above the CoM and act to lift the nose."
Lift and drag are vertical and horizontal components of total aerodynamic force, so I agree you are correct since this is a rotation, but is this a distinction without a difference?

 

[A.T.]

Haha, I absolutely do not fully understand precession.

This video might help to understand the basic mechanism:

I appreciate the responses. How does the spin axis reorient in a forward rotation to the incoming airflow without it being called a torque?

Torque is the rate of angular momentum change, which is different from the axis around which the spin axis rotates:

http://hyperphysics.phy-astr.gsu.edu/hbase/rotv3.html

And the spin axis doesn't do a simple forward rotation, but precesses around the velocity vector, as indicated by the cone in the diagram.

 

[johnlpmark]

That was a very helpful video. I do understand precession better.

Regarding this: "And the spin axis doesn't do a simple forward rotation, but precesses around the velocity vector." I think assumes that the velocity vector stays the same? But the velocity vector changes due to gravity accelerating the bullet down through the medium, and the spin axis follows.

So if the velocity vector does a forward rotation, and the spin axis precesses around the velocity vector, isn't the net movement a forward rotation? I'm not understanding the distinction between a forward rotation as done by precession and a forward rotation as done by torque.

Thank you

 

[A.T.]

Regarding this: "And the spin axis doesn't do a simple forward rotation, but precesses around the velocity vector." I think assumes that the velocity vector stays the same? But the velocity vector changes due to gravity accelerating the bullet down through the medium, and the spin axis follows.

Yes, as the aerodynamic force changes, direction the precession also changes to be around the new aerodynamic force vector.

I'm not understanding the distinction between a forward rotation as done by precession and a forward rotation as done by torque.

"Torque" has a specific meaning in physics. Look at the video with the satellite: There is no torque (called "twisting force" there) around the axis that the orbital plane tilts around.

 

[johnlpmark]

So that answers one question, but shifts to another. How does spin stabilization or an aerodynamic upwards, overturning torque (pitch-up) cause the spin axis to rotate down (pitch-down) without there being another, opposing torque? (Drag force changes velocity but not rotation, meaning that the lift force is the only torque acting on the bullet during its flight that could cause rotation.) As I understand it, precession works at 90 degrees, not 180 degrees?

 

[A.T.]

How does spin stabilization or an aerodynamic upwards, overturning torque (pitch-up) cause the spin axis to rotate down (pitch-down) without there being another, opposing torque?

The instantaneous axis that the spin axis rotates around is perpendicular to both (parallel to the cross product of both): the spin axis and the applied torque. See the satellite animation.

If initially the bullet is only pitched up, the initial instantaneous precession will rotate the spin axis sideways (cross product has no sideways component). But as soon the spin axis is rotated a bit sideways, the cross product will have a sideways component, so the spin axis rotates slightly down too. This happens continuously all the way around the cone pictured in the diagram above.

See also precession of a spinning top:
http://hyperphysics.phy-astr.gsu.edu/hbase/top.html

 

[johnlpmark]

Thanks for the diagram. I think I might be understanding, but please let me know your thoughts.

As the diagram pertains to bullets, what is very important is something not shown which is ΔΦ, or change in the angle between the direction of torque and the spin axis. In all the diagrams I have seen, Φ has been set as a constant for simplification. However in actuality, Φ clearly changes over time.

Is ΔΦ < 0 equivalent to dynamic stability, while ΔΦ > 0 equivalent to dynamic instability? http://ffden-2.phys.uaf.edu/212fall2001_Web_projects/Isaac Rowland/Ballistics/Bulletflight/stab.htm

I think that the question of why bullets differ from typical diagrams of gyroscopes is that in typical diagrams, Φ is 90 degrees (i.e., the spin axis is perpendicular to the direction of torque). Therefore, the spin axis aligns to the direction of torque with both positive and negative ΔΦ. In contrast, with the diagram of the spinning top, only ΔΦ < 0 does the spin axis align.

The difference is that ΔΦ > 0 occurs because gravity creates torque versus ΔΦ > 0 occurs because precession occurs at a speed that redirects that torque at greater rate that it occurs. Does this all sound right?Thanks

 

[A.T.]

In all the diagrams I have seen, Φ has been set as a constant for simplification. However in actuality, Φ clearly changes over time.

It is constant if there is no friction and the force is applied at a fixed point on the spin axis, with a fixed magnitude and direction.

The wooden toy stands on a rounded tip, so the point of force application is not exactly on the spin axis. Similarly for bullets the aerodynamic force application point is also not exactly on the spin axis. You also have dissipative friction and the Magnus effect.

 

[johnlpmark]

Thanks for the clarification on why it occurs. Is the rest of what I wrote an accurate analysis?

 

[A.T.]

ΔΦ < 0 equivalent to dynamic stability, while ΔΦ > 0 equivalent to dynamic instability? http://ffden-2.phys.uaf.edu/212fall2001_Web_projects/Isaac Rowland/Ballistics/Bulletflight/stab.htm

Yes, that is how I understand their definition of static vs. dynamic stability.

When you click on the links on that page you get to the sub-pages, with formulas for the stability-conditions. As you see dynamic stability is quite complex, so I'm not sure if you can capture all that in some simple explanation.

 

[johnlpmark]

So from what I gather, this is the answer to the question. Can you let me know if you agree with it please?

The first concept is the center of mass. The center of mass is the average location of mass of an object. The center of mass does not have to be within the bounds of the object, as in a donut or hollow-point bullet, where the center of mass may be in an interior area that is vacant.

The second concept is the aerodynamic force. For the purposes of this issue, this is the force that is generated on a solid object when it collides with gas. This is conceived of as the solid object traveling through the gas; however it is only the differential in velocity that matters. Whether from the perspective of the observer the solid object moves through a stationary gas or whether a moving gas goes around a stationary, solid object, the same force occurs.

Lift and drag are perpendicular components of the aerodynamic force, and not independent forces. The partial vertical component of the aerodynamic force is lift, which the horizontal component is drag. Specifically, lift is defined as being parallel to gravity while drag is perpendicular to gravity.

The third concept is the center of force. This is the average location of force acting on an object. Special cases of the center of force include the center of gravity (technically gravity is not a force, but here it is treated as one) and the center of pressure. The center of gravity is identical to the center of mass when gravity is the same along an entire object. The center of pressure is the average location of the aerodynamic force on an object. Bullets taper at the front where the air impacts them, therefore bullets have a center of pressure in front of the center of mass.

In the context of ballistics, lift is also called the overturning force. As a bullet travels, the bullet naturally does not point straight into the air; it is always slightly eschew. Therefore, one side of the bullet it hit harder by the air and receives more force. This force differential between the top and bottom, combined with the fact that the center of pressure is in front of the center of mass, leads to an overturning torque on the bullet would cause the bullet to overturn and tumble if not for spin stabilization.

The fourth concept is spin axis. When an object rotates, the axis about which it rotates is called the spin axis. An object wants to rotate around the center of mass; however, they actually rotate around the average center of mass. Rotating around a spin axis creates gyroscopic stability, which is the principle that a spinning object tends to maintain its rotational axis. This stability is a result of the conservation of angular momentum, a fundamental concept in physics. When an object spins, it resists changes to its orientation due to the angular momentum generated by its rotation.

Pulling the previous concepts together, when bullets spin, they rotate their average center of mass around the spin axis. This spinning maintains the center of pressure in front of the center of mass. In doing so, the tapered end of the bullet stays pointed forward, maintaining the bullet in an aerodynamic orientation.

The mechanism by which spinning maintains the center of pressure in front of the center of mass is called gyroscopic precession. Precession is a phenomenon that redirects force (y axis) that is perpendicular to the spin axis (x axis) to be perpendicular to both (z axis). For example, when an upright, clockwise-spinning wheel is suspended in the air by a string on one side and nothing on the other, the force of gravity pulls down, and the wheel rotates to the left.

Curiously, bullets at first would appear to rotate in the opposite direction of the aforementioned wheel. When bullets pitch down, they turn to the right. However, this is not actually a difference in how force is redirected in the horizontal direction, but the vertical direction. First, bullets turn right instead of left because the net overturning torque pushes the tip of the bullet upwards. This is in contrast to the wheel, where the net force of gravity attempts to pull the unsupported side of the wheel down. Upward pitching torque causes a right turn and a downwards pitching torque causes a left turn.

Notably in both cases, without friction and point connections and with infinite angular momentum, precession would be perfect and no vertical movement would occur at all. However in the real world, gyroscopically stabilized objects do still rotate parallel to an applied force. That means there is actually a second spin axis about which the first spin axis rotates. Referring back to the wheel example, although it does rotate left, gravity also still rotates the unsupported side of the wheel down. Curiously though, bullets also pitch down! For the wheel the explanation is simple: the force of gravity pulls the wheel down. But the way that a bullet defies the net overturning torque to actually turn into the torque is explained by a concept called dynamic stability.

Dynamic stability is the ability of a gyroscopically stabilized object to align the spin axis with the force vector. (In more technical, ballistic terms, dynamic stability is when yawing motions and radii of nutation and precession decrease over time.) A good demonstration of dynamic stabilization is a spinning top. Watch any video of a spinning top and no matter the initial orientation of the spin axis, they immediately orient themselves directly upwards and align with the gravitational force vector. Once angular velocity drops so too does dynamic stabilization and the spin axis slowly deviates from the force vector. Spinning wheel and gyroscope demonstrations usually do not demonstrate dynamic stability because they initialize the spin axis too far (often perpendicular) to the force vector for alignment to occur. That is, the direction about which the first spin axis rotate depends on whether precession can overcome the applied torque vector to rotate into the torque vector.