I play table tennis and we can hit balls to make them spin in pretty much every way possible except to spiral about its direction of travel like an American football throw. Of course it could be shot from a rifled gun to spin like this but I wonder if it is even possible for a player with a normal paddle to ever get this result. It's pretty clear that it can't be done from a serve (zero spin), but perhaps it's possible during play to combine some particular incoming spin with a shot imparting some other spin such that the net result is a spiral spin. If it is, how? and if not, why not?

Here are the problem conditions: The ball is coming towards you. You get to choose its speed and axis of rotation (if any), and rotation speed. The only restriction is that the incoming axis of rotation can't already be in the direction of flight. You need to return the ball by hitting it at any desired point on its surface with a force and direction of your choosing. Can you create a hit that results in the outgoing ball spinning strongly about its direction of flight? You can ignore gravity, air resistance, etc.

  • $\begingroup$ Certainly you just have to rotate the paddle around the desired axis when you hit the ball. I guess it will have little effect on the flight path of the ball because there is no coriolis force. When it hits the table there will be an effect though. $\endgroup$
    – Jannick
    Aug 16, 2016 at 8:01
  • $\begingroup$ Don't worry about effects from the table, Coriolis force, gravity, etc. Rotating the paddle might impart a slight twist but not a strong one. Also, I just added the condition that the incoming ball is not already spinning about its direction of travel. $\endgroup$ Aug 16, 2016 at 10:11
  • $\begingroup$ Depends on how fast you spin your paddle e.g. with a motor. The answer is it is theoretically possible by spinning the paddle but this may be unpractical with a human arm and a standard paddle. $\endgroup$
    – Jannick
    Aug 16, 2016 at 13:14
  • $\begingroup$ I'm looking for a solution that a human could conceivably perform. $\endgroup$ Aug 16, 2016 at 19:34
  • 1
    $\begingroup$ 2. I think it might help to edit the question to clarify something about the mechanics of spin production in table tennis. In particular, I believe you can think of the paddle's contact with the ball as essentially instantaneous and the paddle's direction of travel as in a straight line, in the sense that dwell time while the ball is in contact with the paddle is so short that a human cannot feasibly change the direction of travel of the paddle within that time. Thus, answers like "rotate the paddle around the ball, maintaining contact during that entire time" are not humanly possible. $\endgroup$
    – D.W.
    Aug 17, 2016 at 23:00

2 Answers 2


This is called corkscrew spin (alternatively cork spin or drill spin) - ie a spin in which the axis of rotation is along the trajectory of the ball.

Spin is imparted to the ball by hitting it with a glancing blow. The very high friction between the ball and the rubber coating drags the very light ball along the face of the bat. To impart corkscrew spin, the racket must be swung perpendicular to the trajectory of the ball. However, to impart forward momentum the racket must at the same time move parallel to the intended trajectory.

It is impossible to achieve both these goals (corkscrew spin and forward motion) with any slicing stroke, especially if the ball bounces off the bat. But it can be achieved by rolling the bat forward around the ball in a 'corkscrew' motion, in imitation of a bullet being twisted out of a rifle. The limited twisting ability of the human wrist restricts what can be achieved in practice to a compromise between a tight roll and a forward thrust, as seen near the start of this video. Because the thrust is applied parallel to the face of the bat it necessarily introduces spin in a perpendicular direction also.

From what I have seen in this and other online videos, it does seem to be practically impossible either to serve or return a ball with pure corkscrew spin. A small amount of side or back spin seems to be inevitable.

In regard to technique, US certified national coach Larry Hodges explains :

Corkscrew spin is not too common in table tennis, and is usually only used by advanced players when serving. It is difficult to produce except with a high-toss serve (i.e. a serve where the ball is tossed 6-10 feet or more into the air). Sometimes, a player out of position will scoop a ball off the floor, and when the ball hits the table, it jumps sideways because of corkscrewspin. Lobs and counterloops also may have this type of spin.

Judging by the comments in Larry Hodges' coaching forum, it is very difficult to master and takes a lot of experimenting to get it right. A student explains :

I think I came close to hitting one using FH pendulum technique with the racket being kept vertical through the swing. If I timed it so I hit the ball very late in the swing as the tip was coming up on the side of the ball I could get that almost 90 deg sideways jump on the bounce. However, I had lots of trouble keeping the serve low since I was striking the ball with such an upward motion.

Larry replies :

To put corkscrewspin on a serve, you have to really get under the ball. It's much easier to do with a high toss. The ball will then take a big sideways hop when it hits the table.

To keep the ball low, you need to contact the ball very low to the table, with a very fine grazing motion. Your contact point will be mostly under the ball, but towards your side of the ball. You'll have to experiment a lot on this as there's a wide range of contact points, depending on the angle and direction your racket approaches the ball from and how finely you graze the ball.

Note: The last paragraph referred to in a comment by DW has been moved and is now 4th paragraph.

  • 2
    $\begingroup$ That's a good answer and takes it about as far as I've gotten on my own but doesn't seem definitive. Your argument about needing forward momentum is why I convinced myself it's impossible from a static serve. A high serve is different and perhaps enough but has no spin. It feels maybe possible if we get to choose the incoming spin but it's too complex to see. I think a mathematical proof or argument is needed. $\endgroup$ Aug 17, 2016 at 22:29
  • $\begingroup$ Great answer! I think the last paragraph is the key part here. It doesn't seem possible to serve with pure corkscrewspin, because you need to impart some forward momentum. But it does seem possible to serve some with combination of corkscrewspin + sidespin, and it might be possible to make it overwhelmingly corkscrewspin (say 80% corkscrewspin + 20% sidespin). I don't think the answer changes much if you're returning someone else's hit; they'd need an infeasible amount of underspin for you to be able to hit purely under/over the ball and have it reverse direction. $\endgroup$
    – D.W.
    Aug 17, 2016 at 23:35
  • 2
    $\begingroup$ @MelindaGreen , just to clarify, when you say "static serve", you mean hitting the ball when it's at the top of its pure up-then-down trajectory, i.e. when its velocity is zero? If so, yes, a corkscrew spin is impossible from a static serve: in general, the linear velocity and angular velocity imparted by a single impulse are always perpendicular. This follows from the fact that, measured at the object's center of mass, force and torque are perpendicular, so linear/angular momentum imparted are perpendicular, and an impulse is the limit of these things as the time of contact goes to zero. $\endgroup$
    – Don Hatch
    Aug 18, 2016 at 6:47
  • 1
    $\begingroup$ ... and the fact that torque is perpendicular to force can be said to follow from the definition: that is, torque is the cross product of lever arm and force. $\endgroup$
    – Don Hatch
    Aug 18, 2016 at 6:53
  • $\begingroup$ Yes, think of the static case with no gravity or air and a motionless ball. The corkscrew shot is impossible. The question is whether a specially prepared ball not spinning on the desired axis can be hit at a special point of contact with a force vector resulting in the desired flight. $\endgroup$ Aug 18, 2016 at 9:26

If the paddle is allowed to be in contact with the ball for more than an instant, it seems fairly obvious that anything's possible. So we assume an ideal situation in which hits are instantaneous transfers of linear and angular momentum at a single point on the ball's surface, as stated.

Case 0: There's no gravity, and the ball starts motionless.

The initial hit imparts some linear momentum, and some angular momentum perpendicular to the linear momentum. This perpendicularity follows from the fact that force is perpendicular to torque, by definition of torque. So no, it's not possible to get a corkscrew spin (i.e. linear and angular momenta parallel) in a single hit starting from motionlessness.

Here's how to get it in 3 hits, though. Give player 1 a paddle with some amount of friction, and give player 2 a frictionless paddle. Ball starts motionless. Player 1 gives the ball a sideways whack. Ball now has nonzero linear velocity and nonzero angular velocity (perpendicular to the linear velocity, as previously noted, although that doesn't matter for the rest of the argument). Player 2 gives it a whack that exactly cancels the linear velocity (and does nothing to the angular velocity-- player 2 can never change the angular velocity since eir paddle is frictionless). Ball is now spinning in place. Player 2 now gives the ball a second whack, in the direction of (or opposite to) the angular velocity. Ball is now in the desired corkscrew trajectory.

Note that player 2's first whack is actually a "negative" whack-- that is, player 2 actually just lets the ball hit eir paddle while moving the paddle in the same direction as, but slower than, the ball; that's how to cancel linear velocity.

The 3 hits can be improved to 2 as follows: just replace player 2's two hits with a single hit whose linear impulse is the sum of the two linear impulses. Player 2's frictionless paddle makes this straightforward (there are no angular momentum impulses to consider).

Case 1: Constructing a case where the ball starts with a nonzero (say, downward) velocity, with no spin. A real-life serve falls into this case.

In this case, you can achieve a corkscrew spin in a single hit (contradicting the OP's claim that this can't be done on a serve). Here's how to construct such a hit-- it's easiest if I'm allowed to choose the initial velocity based on other results, so I'll take that liberty. First, assume no gravity and the ball starts motionless, as before. Hit the ball sideways, giving it some amount of linear and angular velocity (perpendicular to each other, as previously noted). Let $v_1$ be the linear velocity after the hit. Choose any desired final linear velocity $v_2$ parallel to the angular velocity. Now switch to a reference frame that is moving at constant velocity $v_1 - v_2$ with respect to the original reference frame, and consider what the previously described event looks like from the new reference frame. In the new reference frame's coordinate system, the ball's initial linear velocity was $v_2 - v_1$, and its initial angular velocity was 0; after the hit (which adds $v_1$ to the linear velocity in either coord system), the ball's linear velocity is $v_2$, parallel to its angular velocity, as desired.

Case 2: Constructing a case of nonzero initial velocity and nonzero initial spin that's not around the velocity vector.

This one's easy, but not as interesting or useful as the previous case (it can't happen in a legal serve, since in a legal serve the ball has zero spin when it first hits the paddle).

The simplest way to think about this is to work backwards in time. That is, start with any corkscrew spin of your choosing; it's easy to change the corkscrew spin to a non-corkscrew spin with one hit: just hit the ball, with no friction, in any direction that's not parallel to the direction it's going in; that knocks it out of its corkscrew spin into some non-corkscrew spin (with nonzero linear and angular velocities). Now play the movie backwards, thereby achieving the desired result of changing a non-corkscrew spin to corkscrew spin in one hit.

Update: if we wish to make a case of Case 2 that conforms to the additional constraint that initial and final angular velocities aren't in the same direction (this additional constraint is specified in Melinda's comment below; I don't see it in the original question), just modify the initial (before time reversal) hit so that, instead of being frictionless, it imparts some random spin; probability of that random angular velocity impulse being in the same direction as the original angular velocity is zero, so the final nonzero angular velocity will be in a different direction from the initial nonzero angular velocity. Then play the hit backwards as before.

  • $\begingroup$ I don't follow the bit about the change of coordinates but it sounds like it may be in conflict with my condition that the ball not be prepared as already spinning about the desired direction of travel. You don't need a series of hits to set it up. You may start the ball with any combination of spin axis, spin speed, linear direction and linear speed. All of these are relative to the desired output direction and spin. 'The only restriction is that the chosen initial spin axis can not also be the desired axis. The shot must create the desired axis and the linear speed in that direction. $\endgroup$ Aug 18, 2016 at 9:16
  • $\begingroup$ The change of coordinates isn't in conflict with the condition of no initial spin. The new coordinate system differs from the old in linear velocities only, and the two coord systems are not rotating nor accelerating with respect to each other. Ball's spin (angular velocity, angular momentum) is the same in both coord systems at all times; in particular, the initial spin is zero. $\endgroup$
    – Don Hatch
    Aug 18, 2016 at 9:18
  • $\begingroup$ I added "case 2" for obsessive completeness (since the rules may be interpreted as disallowing case 1's zero initial spin) even though I think case 1 is the interesting one. Not sure whether that was a sticking point or not. $\endgroup$
    – Don Hatch
    Aug 18, 2016 at 11:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.