1 / 27

Physics 199BB The Physics of Baseball

gardenia
Download Presentation

Physics 199BB The Physics of Baseball

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

    1. Week 4 1 Physics 199BB The Physics of Baseball Fall 2007 Freshman Discovery Course Alan M. Nathan 403 Loomis 333-0965 a-nathan@uiuc.edu Week 4

    2. Week 4 2 Forces on a Baseball in Flight Gravity Already discussed Drag (“air resistance”) Force Already discussed Magnus Force Now we do this

    3. Week 4 3 The Magnus Force

    4. Week 4 4 Recall our definitions ? is “angular velocity” a measure of how fast the ball is spinning units are rad/s or rev/min (rpm) to convert from rad/s to rpm multiply by 60/(2?) to convert from rpm to rad/s divide by 60/(2?) ? has a direction

    5. Week 4 5 The spin axis is the line connecting the south to north pole (“right-hand rule”)

    6. Week 4 6 The Magnus Force: The magnitude of FM FM = ˝CL?Av2 CL is the “lift coefficient” CL = CM(R?/v) FM = ˝CM?AR?v CM is the “Magnus coefficient” A dimensionless number

    7. Week 4 7 The Magnus Force: Some numerology rho=1.23 kg/m^3 A=4.16e-3 R=0.0364 9.55 rpm/rad/s 4.45 N/lb 1 mph/0.447 m/s Cd=1/2 All this implies C_M=1.02 Work out on board. Numerical example: V=90 mph W=1800 rpm Adair: FM=0.27 Cd lb If Cd=0.3, FM=0.097 lb Wt = 0.319 lb FM/Wt = 0.30 (see Fig. 2.2, p. 12) If CM=1, then FM=0.159 lb, FM/Wt=0.497rho=1.23 kg/m^3 A=4.16e-3 R=0.0364 9.55 rpm/rad/s 4.45 N/lb 1 mph/0.447 m/s Cd=1/2 All this implies C_M=1.02 Work out on board. Numerical example: V=90 mph W=1800 rpm Adair: FM=0.27 Cd lb If Cd=0.3, FM=0.097 lb Wt = 0.319 lb FM/Wt = 0.30 (see Fig. 2.2, p. 12) If CM=1, then FM=0.159 lb, FM/Wt=0.497

    8. Week 4 8 The Magnus Force: Numerical Examples Numerical example: V=90 mph W=1800 rpm Adair: FM=0.27 Cd lb If Cd=0.3, FM=0.097 lb Wt = 0.319 lb FM/Wt = 0.30 (see Fig. 2.2, p. 12) If CM=1, then FM=0.159 lb, FM/Wt=0.497Numerical example: V=90 mph W=1800 rpm Adair: FM=0.27 Cd lb If Cd=0.3, FM=0.097 lb Wt = 0.319 lb FM/Wt = 0.30 (see Fig. 2.2, p. 12) If CM=1, then FM=0.159 lb, FM/Wt=0.497

    9. Week 4 9 The Magnus Force: The direction of FM Force acts in the direction another right-hand rule force is perpendicular to both v and ? force acts in the direction that the leading edge of the ball is turning

    10. Week 4 10 Some Qualitative Effects of the Magnus Force Backspin makes ball rise “hop” of fastball undercut balls: increased distance, reduced optimum angle of home run Topspin makes ball drop “12-6” curveball topped balls nose-dive Breaking pitches due to spin Cutters, sliders, etc. Golf…Golf…

    11. Week 4 11

    12. Week 4 12

    13. Week 4 13 Incorporating Magnus into Excel We now have a 3-dimension problem For our first examples, we will only consider 2-dimensional problems topspin or backspin, but no sidespin

    14. Week 4 14 Case 1: Backspin FMx = -FM sin(?) FMy = +FM cos(?) aMx = -2.09 x 10-6 CM?vg sin(?) = -2.09 x 10-6 CM?gvy aMy = +2.09 x 10-6 CM?vg cos(?)= +2.09 x 10-6 CM?gvx

    15. Week 4 15 Case 2: Topspin FMx = +FM sin(?) FMy = -FM cos(?) aMx = +2.09 x 10-6 CM?gv sin(?)= +2.09 x 10-6 CM?gvy aMy = -2.09 x 10-6 CM?gv cos(?)= -2.09 x 10-6 CM?gvx

    16. Week 4 16 Now look at the file: trajectory-drag-Magnus-2da.xls Batted balls Low initial angles range increases angle for maximum range decreases trajectory more asymmetric Higher initial angles range decreases trajectory more symmetric cusps and loops

    17. Week 4 17 Now look at the file: trajectory-drag-Magnus-2db.xls Pitched balls Backspin reduces drop (fastball) Topspin increases drop (curveball)

    18. Week 4 18 How do we know what CM is? An Experiment Done At UIUC

    19. Week 4 19 Tracking The Trajectory

    20. Week 4 20

    21. Week 4 21 Typical Data Discuss analysis: Curvature?vertical acceleration >g for topspin, <g for backspin ?lift forceDiscuss analysis: Curvature?vertical acceleration >g for topspin, <g for backspin ?lift force

    22. Week 4 22

    23. Week 4 23 Results for Lift Coefficient CL Remove Adair curve Make better curve for rest of dataRemove Adair curve Make better curve for rest of data

    24. Week 4 24 Results for Drag Coefficient CD

    25. Week 4 25 Data Do Not Agree with Adair Experimental Data: CM ? 1 for S=0.1-0.3 For 2000 rpm, S=0.1-0.3 corresponds to 57-171 mph For 1000 rpm, range is 85 to 255 mph So, most of interesting range is covered Adair (see p. 24) For 2000 rpm CM=0.8 at 50 mph-–agrees with data (0.8) CM=0.4 at 100 mph—too low (1.1) I have written a paper about this (see web site)

    26. Week 4 26 Now let’s include sidespin z is third dimension, points to pitcher’s right Let’s look at pitched ball only Spin axis lies in y-z plane ?=0 for backspin, 180 for topspin ?=90 or 270 for pure sidespin

    27. Week 4 27 Here are the formulas FMx = FM {sin (?)vz/v-cos(?)vy/v} FMy = FM cos (?)vx/v FMz = -FM sin(?)vx/v aMx=2.09x10-6 CM?g {sin(?)vz-cos(?)vy} aMy=2.09x10-6 CM?gcos(?)vx aMz=-2.09x10-6 CM?gsin(?)vx Notes: when ? is 0 or 180, these formulas are identical to the ones previously used Since v?vx, FMx?0 FMy is responsible for up/down break (max when ?=0 or 180) FMz is responsible for left/right break (max when ?=90 or 270) 5. FM makes angle ?+90 with z axis

More Related