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Can a rising fastball exist?

  1. Apr 2, 2009 #1
    Hi, I was wondering if a rising fastball can't exist because no pitcher can generate the necessary rpm's to create a lift force greater then the baseball's weight or are there rules of physics that prevent it from ever happening.

    Thanks
     
  2. jcsd
  3. Apr 2, 2009 #2

    russ_watters

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    The answer to the question in the title is yes, but you need to be very clear about what you mean and you weren't. There are several scenarios where it does and several where it doesn't exist:

    1. There is currently a sidearm pitcher in MLB who'se release point is ankle-high and therefore his fastball rises as it goes toward the batter's box.

    2. A typical overhand fasball can't rise relative to the pitcher's hand because the release point is above the batter's box.

    3. A typical overhand fastball can't drop toward the ground, then rise toward the batter's box because the angles are too steep and the lift insufficient.

    4. To a batter's eyes, a "rising fasball" as typically described, does not arrive at the batter's box higher than when it left the pitcher's hand because of #2 and 3. What it does do is rise relative to where it would have gone had it not been spinning. In other words, to a batter's eyes, it takes a higher path than it should: it appears to rise.
     
  4. Apr 2, 2009 #3
    Well, I am doing a wind tunnel experiment with a fastball. I was wondering if I cranked the rpm's enough on the fastball (3000 rpm-4000 rpm) would I see a lift force on the baseball greater then the weight of it. Even though, I know it is physically impossible for a human being to generate that much spin on the ball.
     
  5. Apr 2, 2009 #4
    I'd be interested in hearing about your results. A scuffed and unscuffed ball could yield different values.
     
  6. Apr 3, 2009 #5
    pitcher's like Nolan Ryan would throw a fastball so hard and when they let go of it would start to spin backwards and rise when it came in towards the plate.
     
  7. Apr 3, 2009 #6

    rcgldr

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    The real question would be if a human can throw a baseball so that the upwards acceleration due to Magnus effect is greater than the downwards acceleration due to gravity. From all the articles I've read this isn't humanly possible with a baseball or softball. It's easily done with a ping pong ball though.
     
  8. Apr 3, 2009 #7
    It seems there should be an upper limit on lift, no matter how fast the ball is spinning, Jeff.
    I just began reading about the question http://home.iitk.ac.in/~tksen/TKsite/RobinsMagnus/Robins_Magnus.pdf" [Broken]. It's a bit difficult to read, though.
     
    Last edited by a moderator: May 4, 2017
  9. Apr 3, 2009 #8
    what is the difference between the Magnus effect and Bernoulli’s principle.
     
  10. Apr 3, 2009 #9

    rcgldr

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    Magnus Effect - an spinning ball moving at a forward speed will experience an aerodyanmic force perpendicular to it's direction of travel, causing it to curve. Most articles state that this is due to the ealier detachment of flow on the "upwind" surface versus the "downwind" surface, which deflects (accelerates) the air, which in turn reacts with a component of force perpendicular to the direction of travel.

    Bernoulli principle - In an ideal flowing fluid or gas, it's theoretically possible that an interaction between a solid, and that fluid or gas can result in an exchange of kinetic energy for pressure energy or vice versa, with no net work done on that fluid or gas. Since the total energy is constant, then a formula can be created based on constant mass flow. The typical example is an ideal fluid (incompressable, zero viscosity) flowing horizontally (no effect from gravity) through a frictionless pipe of varying diameters. Since mass flow is constant, the fluid through the narrower sections of the pipe has higher velocity, inversely proportional to cross section area. The pipe doesn't perform any work on the fluid, so the only remaining cause for the increase in velocity is an internal pressure differential in the fluid. The pressure in the narrower, faster flowing sections must be less than the larger slower flowing sections.

    The idea here is that since temperature and mass of the fluid is constant, then the average velocity of the molecules also remains constant. Pressure is due to collisions between molecules based on their random velocities. If there's a net flow change without any change in total energy (no work done), then the molecules velocities are more organized and less random, so there are fewer collisions and less pressure. It's important to note that this is requires a change in flow with no work done. If it's internal with no work done, then you get the Bernoulli effect. If the change is due to work done, then the total energy is changed, and during the time work is being done, Bernoulli effect doesn't apply.

    In real life, some work is done when causing any change in flow, but if the work done is relatively small, then the reaction is Bernoulli "like". For example, in the case of a propeller, Bernoulli applies to the flow fore and aft of the propeller, but not across the virtual disk formed by the rotating propeller, because the mechanical interaction of the propeller, causes a pressure jump (and then the higher than ambient pressure continues to accelerate aft of the prop as it's pressure deacreases).

    http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

    Not explained in the nasa link text is fact that the cross section of the air flow aft of the prop is decreasing in radius (otherwise mass flow wouldn't be constant), although the diagram shows it happening.

    Rather than try to deal with the total energy of a fluid or gas, the Bernoulli equation is based on a concept called total pressure, which is the sum of it's static pressure (random) and dynamic pressure (organized). The dynamic pressure = 1/2 x density x velocity^2, and represends the pressure obtained by declerating the fluid's velocity to zero (relative to some frame of reference). Note that static and dynamic pressures are defined as a energy per unit volume = force / unit_area, and not per unit mass. At sea level, pressure is 2116.8 pounds / foot^2. For a volume of 1 cubic foot, the static pressure energy = 2116.8 pounds / foot^2 x 1 foot^3 = 2116.8 foot pounds, an energy unit.
     
    Last edited: Apr 3, 2009
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