The General Description of the Tennis Strokes. For math oriented players only!

Discussion in 'Tennis Tips/Instruction' started by toly, Aug 14, 2012.

  1. toly

    toly Hall of Fame

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    Anatoly Antipin

    1. The General Description of the Tennis Strokes

    The tennis is the game about movement. There is a lot of body activity, a racquet and ball motions etc. It is the common practice to use Vectors to describe any object motion.

    Definition: The vector is a straight line segment whose length is magnitude and whose orientation in space is direction.

    Mostly, I will examine a velocity of the different objects.

    Definition: The velocity is the vector. Its magnitude defined as the speed and orientation in space is the direction of the moving object.

    All these vectors stuff maybe is not really very important, but can be helpful for understanding different tennis techniques. I’ll try to keep the matter as simple as possible.

    The main distinction among the different tennis strokes

    Definition: Flat stroke – the ball is not rotating after the stroke.

    Definition: Spin stroke (topspin, sidespin, backspin, etc) - the ball is rotating after the stroke.

    Question: How can tennis player produce the strokes according to their definitions?

    Answer: If and only if the velocity of the racquet VR (Figure. 1.1) is the perpendicular to the strings plane around impact spot, the stroke is flat. Or, in other words, if the racquet is moved in the direction of the perpendicular to the strings plane around impact spot, the shot will be flat.

    [​IMG]

    Figure 1.1. Racquet velocity (VR) in the case of the flat shot. Coordinate axis OZ and vector VR are perpendicular to the strings plane

    In all other cases, there would be spin strokes. Since, it is practically impossible to maintain the velocity of the racquet perpendicular to the strings plane, in reality, there are no pure flat strokes.

    What happens if the velocity of the racquet VR (Figure. 1.2) has an arbitrary direction?

    Accordingly to the linear algebra, every vector of the racquet velocity VR can be decomposed as a sum of two orthogonal (perpendicular to each other) components (vectors). The flat (or normal) component VF is the perpendicular to the strings plane and the spin (or tangential) component VS is the parallel to it.

    [​IMG]


    Figure 1.2.The arbitrary vector of the racquet velocity (VR) along with its components (VF, VS). The strings plane exists in the XOY plane

    The flat component is mostly responsible for the boll velocity after impact VB (its direction and speed).

    The spin component provides the ball rotation, but with some strings interference can also change the velocity of the ball VB. It can happen due to the racquet strings could “bite” the ball or provide enough friction between the ball and strings and move the ball in direction of the spin component.

    In case of the flat stroke, the spin component usually is equal to zero.

    Something else about spin component

    Since, the spin component VS is the vector, it can be also decomposed on two orthogonal components in the strings plane.

    I assume that racquet stings bed is perpendicular to a ground. Then, the vertical spin component VSpinVer resides in plane perpendicular to the court’s ground. The horizontal spin component VSpinHor belongs to the plane parallel to the court’s ground (Figure 1.3).

    [​IMG]


    Figure 1.3. Vector of the spin component along with its components

    Relationship between defined vectors and common tennis slang

    The upward vertical spin component VSpinVer brushes the ball up, makes topspin, and also can drag the ball upward.
    The downward vertical spin component brushes the ball down, produces backspin, and pushes the ball down.
    The horizontal spin component VSpinHoz makes right/ left sidespin and moves the ball to the right/left.

    In general, the boll spins in the direction of the spin component VS (brushing ball direction) and moves in direction of the flat component VF (perpendicular to the strings plane).

    Digging effect

    Sometimes the spin component cannot produce any significant ball rotation but can change considerably the ball direction and speed.

    For instance, if the point of contact is the sweet spot and the flat component VF has very high speed. The ball just digs into strings bed; the strings hug the ball very hard and do not allow creating any ball’s rotation during “digging” phase of the impact.
    Then strings catapult the ball very fast and the spin component does not have enough time to rotate the ball.

    However, while the ball is in digging and catapult phases the spin component still moves the racquet in its direction and as a result it varies the ball’s direction and speed.
     
    Last edited: Sep 15, 2012
    #1
  2. sureshs

    sureshs Bionic Poster

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    In Fig 1.2, VR cannot be decomposed into VF and VS perpendicular and parallel to the string plane unless VR lies in a plane perpendicular to the string plane.
     
    #2
  3. user92626

    user92626 Legend

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    WTH, I thought people played sports to get away from classroom. Nerds ruin everything!!! :)
     
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  4. toly

    toly Hall of Fame

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    Two thin orthogonal lines define the plane perpendicular to the string bed. The arbitrary vector VR lies in that plane. Think a little bit please!:)
     
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  5. toly

    toly Hall of Fame

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    Sometimes we need a strong jolt!:evil::)
     
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  6. sureshs

    sureshs Bionic Poster

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    That is what I am saying. If VR lies in that plane perpendicular to the strings, it is not an arbitrary vector, so it cannot be used to describe general situations.
     
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  7. toly

    toly Hall of Fame

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    What about red vector?

    [​IMG]

    There is nothing special regarding it. It is also arbitrary vector. Give me any vector’s coordinates and I construct orthogonal plane to the string bed and vector will lie in this plane. How do you think I draw these vectors? I don’t use any extraordinary methods. :confused::)
     
    Last edited: Aug 14, 2012
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  8. sureshs

    sureshs Bionic Poster

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    Yes got it now. The way you drew it, VF and VS cannot be as shown, if VR is going up, away and towards the tip of the racket.
     
    #8
  9. soyelmocano

    soyelmocano Rookie

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    I've often imagined tennis and golf using images similar to this in my head. In my mind you have the initial velocity and launch angle. Of course the speed is mostly determined by how fast the racquet head is moving. Launch angle came be the same either with open orclosed face (and whatever angle in between). For example, to achieve a launch angle of 45 deg upwards, simply angle the racquet 45deg and swing straight along that path. However, one could get the same launch angle with a square face. Now a faster swing on a steeper plane is required to overcome the natural deflection angle.
    Once the ball leaves the strings I imagine the denser cushion of air on the leading spin side pushing the ball in the opposite direction, and gravity pulling down.
    I taught myself how to hit a draw in golf laying in bed at night. I had tried a lot of hitting, but it didn't work until I imagined and understood what was happening.
    It really helps to understand the principles of physics to visualize shots should be. However, once you're on the court, just let it flow.
     
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  10. FrisbeeFool

    FrisbeeFool Rookie

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    Toly, I majored in math in college, and I disagree with pretty much every comment you've made about tennis strokes. Most of your comments seem to be about how players are wristing their grounstrokes, and vectors can prove it somehow. I think If anything is happening with the wrist at the end of the stroke, it's because of the preparation that happened earlier. In my mind, players have relaxed, fluid strokes, and their wrists are relaxed at the end. I'm not sure how vectors fit into all your wristy groundstroke arguments.

    I question whether you understand even basics concepts surrounding vectors. Most of your posts on technique are so off the wall, then you will draw a vector and claim it somehow supports your post. I'm not seeing the connections.
     
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  11. Uthree

    Uthree Rookie

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    If you want to do a 3D analysis, probably more value in describing the players movement using the the 3 axis. Players tend to do to little and/or too much rotation around X,Y,Z axis. This would be a great way to understand the shot but doesn't seem to get much airplay around here.
     
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  12. toly

    toly Hall of Fame

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    Thank you very much. I really appreciate any negative comments, but can you be more specific please? For example, what is wrong with vectors? IMO, they give us good visual representation of motions. Maybe you know something better?:confused:
     
    Last edited: Aug 16, 2012
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  13. toly

    toly Hall of Fame

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    I’m talking about 3D, see please post 1, fig.1.1 and fig1.2.
    I simplified fig. 1.3 (2D), because I don’t want to upset people too much.:)
     
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  14. toly

    toly Hall of Fame

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    Practically, there is no deflection angle in tennis for the reason that racquet strings work as catapult.:)
     
    Last edited: Aug 19, 2012
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  15. sureshs

    sureshs Bionic Poster

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    End of thread? This is all there is to the general description of tennis strokes?
     
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  16. toly

    toly Hall of Fame

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    Yes, this is the end of thread, maybe because everything is extremely simple and tremendously clear.:):confused:
     
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  17. TheTsongaKid

    TheTsongaKid Banned

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    What are the relative percentages of Vf and Vs to maximize a heavy topspin ball with good pace as well?
     
    #17
  18. corners

    corners Legend

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    If the spin component of the swing is there, there will probably always be spin as a result. Please see this paper by R. Cross and take note of

    2. Spin Generation
    5. Racquet rotation
     
    #18
  19. Funbun

    Funbun Professional

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    I have to agree here. You made tennis look unnecessarily complicated here, toly.

    For an observational standpoint, I think you'll have to incorporate much more than just vectors here. If a standard player were to look deep and study your post, they wouldn't get around to hitting very well without more information.

    There's a lot more to tennis than just vectors. You have to bring physics into this. Why not discuss force? Or leverage? Momentum? Vectors only go so far into explaining something literally everybody on this forum does when hitting a tennis stroke.

    The biggest problem I have with this post is that it focuses too much on the racquet itself. The player is the one who controls the racquet, and therefore hits the ball. How about you post something that discusses how to attain the maximum magnitudes for the vectors you described in your original post?

    Taking a solely math-based approach to tennis is limiting. It's better if you expanded to physics instead; it'll be much more applicable. Tennis is a physics-based game, anyway.
     
    Last edited: Aug 20, 2012
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  20. toly

    toly Hall of Fame

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    Rod Cross in article, “Physics of the Tennis Kick Serve”, http://twu.tennis-warehouse.com/learning_center/kickserve.php ,described procedure how to calculate ball’s spin.

    [​IMG]

    [​IMG]

    The racquet velocity (VR) and its components (VF, VS).

    “The amount of topspin is shown in Fig. 6 with the symbol S. Experiments and theoretical estimates both indicate that S is given to a good approximation by

    S = 1.45VA

    where S is the spin in rpm, V is the racquet head speed in mph and A is the approach angle in degrees.

    For example, if A= 0 then S = 0 meaning that there is no spin generated at all. If V = 100 mph and A = 30 degrees then S = 4350 rpm. The amount of spin therefore increases with both the speed of the racquet head and the approach angle of the racquet head. The amount of spin also depends on the speed of the incoming ball.”
     
    Last edited: Aug 20, 2012
    #20
  21. pvaudio

    pvaudio Legend

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    While your descriptions from a math standpoint are indeed correct, they only really apply to the classical definition of strokes. They don't apply to modern stroke mechanics. As a simple example, for both flat and "spin" shots, the racquet velocity should be perpendicular to the stringbed by your definition. If you hit a shot with underspin, as in a slice, this doesn't hold. Here, the racquet face is slightly open, so the strings are pointing slightly upward. The racquet velocity will be from high to low, so at impact, the racquet's velocity vector most certainly can be perpendicular to the string bed's normal vector. How? Easy, the ball is not going straight across the net, it needs elevation as well. I think what you meant to say is that for a flat ball, the racquet face must be normal to the direction of the racquet's travel through prior to and after impact. Because as I just pointed out, at impact, you can get any sort of spin you want and have the stringbed's normal vector be normal to the velocity vector.
     
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  22. toly

    toly Hall of Fame

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    Thanks for your comments!
    The problem is that I'm just afraid to talk about physics in this forum. Even with the simplest math I have more than enough complaints. Nevertheless, next time I’m going to talk about physics of FH. :)
     
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  23. toly

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    Again, thanks for very good article. Basically you are right, but IMO significant “Digging effect” can considerably decrease spin and Ross very simple formula S = 1.45VA shouldn’t work in this case. But I have no proof. It is just hunch.:confused:
     
    #23
  24. toly

    toly Hall of Fame

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    The racquet velocity and orientation doesn’t know anything about modern or classical definition of strokes. This math stuff is working in any sports and motions.

    When you hit any stroke you better clearly understand what motion of your body creates flat component VRF – translational motion of the ball (ball’s speed) and how you are going to build VRS – particular spin at the moment of contact.

    For example, in case of the serve VRF can be constructed by using:
    1. Arm counterclockwise rotation – arm pronation
    2. Wrist flexion
    3. Trunk particular rotations and etc.

    The serve VRS can be made by using:
    1. Wrist ulnar deviation
    2. Elbow flexion and so on.

    Then you also should pay attention to the racquet orientation at the moment of contact.

    Unfortunately, people usually learn it by trial and error method or copy somebody’s stroke.:(
     
    #24
  25. soyelmocano

    soyelmocano Rookie

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    I would have to disagree that angle of deflection is not relevant. Imagine a racquet face that is 15 degrees open. Now try to "catapult" topspin with that. Contrarily, with a face that is 15 deg. closed, an upward swing where the strings grab and throw the ball in a initial upward direction if the speed/force of the upward swing are sufficient to overcome the downward angle.
    You have lots of angles and forces either working against each other or working togetherto form each type of shot.
     
    #25
  26. toly

    toly Hall of Fame

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    There is some misunderstanding. I was talking about the law of reflection.

    The law of reflection states that the angle of incidence ϴi of wave reflecting from boundary is equal to the angle of reflection ϴr. So,

    ϴi = ϴr.

    In tennis, the angle of incidence ϴi of incoming ball reflecting from string bed is not equal to the angle of reflection ϴr and approximately

    ϴr=0°. The law of reflection doesn’t work due to “Digging effect”.

    [​IMG]

    Next picture supports that. I copied frames from http://twu.tennis-warehouse.com/learning_center/spinandstiffness.php.

    [​IMG]
     
    Last edited: Oct 16, 2012
    #26
  27. sureshs

    sureshs Bionic Poster

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    I think it will obey the law approximately if frame was stationary. It is the forward movement of the frame which forces the issue.
     
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  28. toly

    toly Hall of Fame

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    The physics of the collision is exactly the same if the racquet is at rest and the ball approaches at 50 mph or ball is at rest and the racquet approaches at 50 mph. The relative speed between them is relevant.
     
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  29. sureshs

    sureshs Bionic Poster

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    It cannot be the same. The initial momentum will be MV in one case and mv in the other, so if momentum is conserved, the final values will be different.

    If I run into a wall, I will be hurt, but much less than if I stood still and the wall ran into me.
     
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  30. toly

    toly Hall of Fame

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    No, it can be and must be the same. It is the absolute truth.

    There are zillions of articles in which authors use this well known fact. See for instance citation from article, “Elastic and Inelastic Collision in Two Dimensions”, http://www.plasmaphysics.org.uk/collision2d.htm.

    "For the sake of simplicity it has been assumed here that mass 2 is initially resting i.e. vx,2=0, vy,2=0 and vx,2'=Δvx,2' (this does not affect the general validity of the final result as the assumption will be dropped again later by referring the velocities to mass 2 explicitly)."

    Don’t try to invent new physics please.:)
     
    #30
  31. sureshs

    sureshs Bionic Poster

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    I still don't get the point. If the ball is moving toward the racket at an angle, and the racket is moving forward towards the intended target area, the ball will not be reflected according to the laws of reflection, but will be reflected much less. Otherwise, you could never return a cross court topspin back crosscourt.

    Conservation of momentum does not even apply here, because there is a net force being imparted to the racket.
     
    #31
  32. sureshs

    sureshs Bionic Poster

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    Mass 2 is resting sets one of the variables to 0. That is not what you are saying. You are saying relative velocities are all that matter. That is like replacing vx1 by vx1-vx2.

    Say you have a small mass m moving horizontally to the right with speed v and colliding completely inelastically with a larger mass M. The two together will then move to the right with speed mv/(m+M). If m was stationary and M moved to the left with speed v, the two will move to the left with speed Mv/(m+M).
     
    #32
  33. FrisbeeFool

    FrisbeeFool Rookie

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    Bingo! No one is going to be able to model the complex things that are going on in a tennis stroke using high school physics in an online forum.

    Just drawing a vector or a picture diagram does not mean you have accurately modeled a complex real life phenomenon.
     
    Last edited: Aug 22, 2012
    #33
  34. pvaudio

    pvaudio Legend

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    Hence why I often say "please don't use math and science where they don't belong".
     
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  35. r2473

    r2473 Legend

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    #35
  36. toly

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    Let’s analyze collision of the resting ball with moving racquet.

    The ball during impact stays on the racquet's string bed just about T = 4 msec. Assume, it leaves the racquet with speed equal to Vb=100 mph (160 km/h). The mass of the ball is mb = 57 grams.

    The ball kinetic energy before impact = 0.The ball kinetic energy after impact Eb =0.5*mb*Vb ^2.

    In physics, power is the rate at which energy is converted. Let’s calculate power P which produced the conversion of the ball kinetic energy from zero to Eb.

    It is P = Eb /T = 20hp.
    Where, hp is horsepower.

    The 20 horses are working very hard during 4 msec to provide the conversion of the ball’s kinetic energy.

    Should you apply during impact your own additional force to help them to do this work? Even if you add your maximum force during collision the result would be negligible. We have to collect kinetic energy during our forward swing. That’s why there is no short swing in professional tennis when pros hit hard strokes.

    So, when we scrutinize real collision of racquet and ball we can ignore the additional force completely!!!:):confused:
     
    #36
  37. bhupaes

    bhupaes Professional

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    Toly is right. The relative speed between the racquet and the ball will determine what happens in the collision. You can look at it from the ball's frame of reference, or the racquet's frame of reference, or a stationary frame of reference for purposes of calculation, and transform the results from one frame to another using simple calculations.
     
    #37
  38. FrisbeeFool

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    Toly, you're missing the point. Your trying to use a simplistic model that's designed to describe the collision of 2 billiard balls or similar situations. In this situation, there is a force being exerted by the person on the racket before, during, and after impact. This force is constantly changing with respect to time, because racket head acceleration is not a constant. In order to understand the movement of the racket or ball, a mathematician or physicist would probably need to use a differential equation or some other more sophisticated method to describe the aggregation of forces on the racket coming from the collision with the ball and the force being exerted by the person(which will be continuing to act on the racket through and after impact.)
     
    Last edited: Aug 22, 2012
    #38
  39. FrisbeeFool

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    You could do that, but the relative speed between the racket and ball would be constantly changing because racket head acceleration is not a constant, due to the variable force being exerted by the hitter with respect to time. You couldn't calculate it using algebra, you'd need to use vector calculus, a differential equation, or some other sophisticated technique that would more accurately model the situation.
     
    Last edited: Aug 22, 2012
    #39
  40. sureshs

    sureshs Bionic Poster

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    That is what I always thought and have done problems with it, till I came up with the example above (inelastic collision) when I was trying to see if toly was wrong. Can you tell me the fallacy in my argument (#32)?
     
    Last edited: Aug 23, 2012
    #40
  41. sureshs

    sureshs Bionic Poster

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    Do you think the racket moved by itself under gravity and its horizontal component was enough to keep it moving to strike the ball? No. The person is exerting a force on the racket.
     
    #41
  42. toly

    toly Hall of Fame

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    The law of conservation of momentum

    If the velocities of the particles are U1 and U2 before the interaction, and afterwards they are V1 and V2, then

    M1U1 + M2U2 = M1V1 + M2V2 (1)

    This law holds no matter how complicated the force is between particles.

    M1 and M2 are constants, but U1, U2, V1, and V2 are not. They depend on an arbitrarily chosen coordinate system. If we change coordinate system, we have to rewrite equation (1), but physics of collision process would be the same.:)
     
    #42
  43. toly

    toly Hall of Fame

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    Yes, you can and maybe should exert force on the racquet to keep proper racquet orientation, but this force practically doesn’t affect collision, because 20 horses are much, much … stronger than you. :)
     
    #43
  44. FrisbeeFool

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    Toly. What is your math background? You're using modeling techniques from an introductory algebra based physics class. These models are designed to introduce basic concepts of physics to someone with no math background.

    Have you taken any physics courses that use modeling techniques from calculus? Have you taken any engineering courses? Are you self-taught?

    The situation you are attempting to describe is complex and has many variables. If you relate these variables using simplified algebra, you won't be accurately modeling the situation.

    We haven't even gotten into wind/air resistance. It would be a huge factor in this situation. I still have my notes from the first week of my ordinary differential equations class. If you want I can scan them and post them. They model a simplified view of wind resistance, in order to introduce the topic. There are many ways to model air resistance with increasing complexity and sophistication. Since I only studied math at the undergraduate level, I don't know them. I'm guessing you don't either.

    Quit wasting people's time. This thread isn't for math oriented players at all. You're trying to use pre-algebra and algebra to model a dynamic, constantly changing situation. Hmmm, what discipline studies rates of change???? Calculus

    The two objects you're examining aren't moving in space uneffected by other forces. The have constant mass, but they don't have constant velocity. They don't have constant acceleration!! Racket head acceleration is not constant!! It's not like two billiard balls with no outside forces at the time of the collision. The force exerted by the person on the racket is effecting the collision.

    I can probably scan old notes from my differential equations courses if you want. Just go to google look up differential equations, and look up simplified models of motion. You're trying to use introductory techniques to describe a dynamic event? It won't work!
     
    #44
  45. sureshs

    sureshs Bionic Poster

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    I really don't follow. You seem to be saying gravity is doing all the work.

    In your example of ball leaving the racket at 100 mph, it could also be a serve, not a forehand. Are you saying that gravity is what provides the force on the serve when in reality the racquet is almost moving up or slightly down at contact? If I just hang the racket and let it fall on the ball toss, a 100 mph serve will be produced?

    Doesn't seem right to me. If the above is not right, then the argument that there is no force exerted by the player on the forehand is also wrong.
     
    #45
  46. sureshs

    sureshs Bionic Poster

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    Do a simple experiment. Hang a racket with a thin string and also hang a ball in front of it. Take the racket back like a pendulum to the same distance that Fed takes back his racket, and let go, and position the ball so that it gets hit.

    See if the ball leaves at 80 mph.
     
    #46
  47. sureshs

    sureshs Bionic Poster

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    That is OK, but I wanted someone to point out what was wrong in my argument
     
    #47
  48. toly

    toly Hall of Fame

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    I have several degrees - in math, radio electronics, computer science, and radio physics. I worked as one of the leader designer of radio locators for more than 20 years. I had few master and PhD students who tried to complicate any assigned work, unfortunately for them they didn’t get job. If you want to contribute some sophisticated and useful analysis to this thread, you are welcome.:)
     
    Last edited: Aug 23, 2012
    #48
  49. FrisbeeFool

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    My bad, sounds like you know your stuff. I guess you're just introducing the algebraic approach to start things off?? If that's really your background, you have to realize racket head acceleration won't be constant? Why don't you explain the different types of collisions and the different models?
     
    #49
  50. bhupaes

    bhupaes Professional

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    I believe the objective is to model the collision in a very small timeframe that brackets the collision. How the racquet and ball achieved their speeds and direction would be immaterial for the purposes of this exercise. I believe this is what toly is trying to do, in order to simplify the model/calculations used, and I don't see anything wrong with it...
     
    #50

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