Hitting the cutoff man (June 13, 2003)
fracas discusses the difference between a relay throw, and throwing it in directly. Can anyone add to his story regarding the competition between the cannon-armed OF, against the relay race? Seems to me that there must be some measurable point at which say a throw from say 223 feet is the break-even for an average throwing fielder, and maybe more for the stronger throwing ones. And of course, the relay has the benefit of the runners moving up.
--posted by TangoTiger at 02:41 PM EDT
Posted 7:33 a.m.,
June 16, 2003
(#1) -
Sylvain(e-mail)
(homepage)
Ok, I'll take my shot on this issue.
Suppositions:
- 3 players considered (OF, IF, C); all 3 are aligned, the infielder is between the outfielder and the catcher.
- spin and magnus force negliged
- drag negliged (for this model)
Initial conditions at t=0 : we suppose the following coordinates:
OF (x=0; y=0), IF (L1; 0), C (L2; 0), the OF has the ball, and the initial vector speed Vo makes with Ox an angle alpha (noted a).
The methodology is not that long and already explained in the homepage link, so I'll begin with the results:
x = Vo*cos(a)*t
z = -1/2*g*t^2 + Vo*sin(a)*t
I also pose a belongs to ]0; Pi/2[ (z has to be >= 0 and x growing, the level z = 0 corresponds to the "altitude" of the shoulder. Generally speaking, the ball will reach the following player (IF/C; coordinates (x = L, z = 0) at a time T if:
z = 0 = -1/2*g*T^2 + Vo*sin(a)*T (1) and x = Vo*cos(a)*T = L (2)
First result : the maximum distance (without rebound) that can be reached by a player that throws a ball with a speed Vo is: Lmax = Vo^2/g. This maximum distance will be reached if a = Pi/4 or a = 45°.
We now suppose that every distance will be in accordance with the condition LIF) = Vo(OF)*sin(1/2*arcsin(g*L1/Vo(OF)^2))*2/g
+ time for the infielder to catch and throw
+ T(IF->C) = Vo(IF)*sin(1/2*arcsin(g*L2/Vo(IF)^2))*2/g
Second Result : It will be faster to relay if the sum of the 3 terms above is bigger than the time needed to throw directly .
But:
- for the direct throw, we need one very accurate throw (because y, the deviation will equal to Voy*time of flight, and Voy will be the measure of the accuracy)
- for the relay, we need 2 accurate throws; the accuracy of the first throw will also influence the time neede by the IF to catch and throw.
Numerical Application
Max distance
Remember: the drag is not considered! The real distances will be shorter (I don't know by how much, but I'd say mimimum 10%)!
Original Velocity (MPH)__Max Distance (FT)
89__536
84__471
78__410
73__354
67__301
Relay or Direct ?
I suppose that both the IF and OF can throw at the same speed (75 mpH)and that the IF is midway between the C and the OF
Distance(OF-C, in FT)___Time Direct(Sec)___Time Relay(Sec)
(the max distance for a speed of 75 mpH is 376 feet)
360__4,1__3,4 + time for the IF to catch and throw
330__3,5__3,1 + ...
300__3,1__2,8 + ...
270__2,7__2,5 + ...
The break even (considering an IF catching + throwing time of 0,4 sec) is at 330 feet distance between C and OF.
I once read the max distance ever thrown at was 180 meters or 590 fee. This corresponds to a speed of 94 mph not considering drag and supposing the original angle was 45°. The real speed might have reached 100 mph.
Sylvain
Posted 9:54 a.m.,
June 16, 2003
(#2) -
tangotiger
Sylvain,
I know you said drag is not considered, but that has to be a key consideration right? I mean, I'm sure Tim Raines can throw a ball at 75mph, but he'd never be able to throw it 330 feet, certainly not on the fly. I'd also think that maybe a throw on 1 or 2 bounces would be better, in terms of "accuracy".
Thanks for producing your work, as it is very fun to try to decipher!
Posted 10:49 a.m.,
June 16, 2003
(#3) -
Sylvain(e-mail)
Tango,
The problem I have with drag is that integrating it gives some very heavy equations that can't be solve easily. I haven't had time yet to really give a look at it, and every time I found a paper on the web it implied (if my memory is correct) a computed solution using maple/mathematica or another software in case of a non vertical movement. Even vertical, still if my memory is correct, the solutions are functions like cosh, sinh.. (cosh = 1/2*(exp(x)+exp(-x), sinh = 1/2*(exp(x)-exp(-x)).
The differential equations actually look like
dx/dt = x*rootsquare(x^2+y^2) and
dy/dt = constant + y*rootsquare(x^2+y^2)
So, if there's any differential equations specialist here, don't hesitate to contact me.
Still, based on the homepage link, tim raines could throw at about 243 feet (angle a = 35°); give me 20 minutes and based on this site I'll try to build a basic rule linking my results to the "real" ones.
As for the deciphering, I'll try to find the time to type everything and make a .pdf out of it. I know it's hard to read.
Sylvain
Posted 11:15 a.m.,
June 16, 2003
(#4) -
tangotiger
No need to go to the trouble for the pdf file. I meant deciphering not in your presentation layout, but in the equation itself. I'm trying to remember my physics classes from 15 years ago.
Posted 11:17 a.m.,
June 16, 2003
(#5) -
Sylvain(e-mail)
So, based on 15 points (small sample size...) between 101 and 67 mph, a basic rule seems to be:
Max Distance (feet) = 3.4 * speed (mph)
Or
Max Distance (feet) = -119.5 + 4,481 * speed (mph)
The angle is always around 35°
SC
Posted 5:52 a.m.,
June 17, 2003
(#6) -
Sylvain(e-mail)
I gotta make such an idiot: here is the homepage link I refer to in post#3:
http://www.npl.uiuc.edu/~a-nathan/pob/
the sim I used to "generate" the 2 formulas:
http://www.scri.fsu.edu/~jac/Java/baseball.html
S
Posted 9:09 a.m.,
June 17, 2003
(#7) -
Sylvain(e-mail)
(homepage)
So, some better results.
Methodology
Based on the equations (and the notations) given in the homepage link, and neglecting the lateral magnus force (phi = 0), I ran a step method (constant step = 0.01 second):
vx(step n+1) - vx (step n) = - vx(step n)* step * Drag_Coefficient(vx(step n);vz(step n)) - step * B * rotation * vz(step n)
B is a constant and corresponds to the magnus force; the drag coefficient depends on the speed, and both formulas and values given already include m.
Add g (gravity) and do the same along the z axis and you will obtain the equation for vz(step n+1).
For x and z:
x(step n+1) = x (step n) + step * 1/2 * (v(step n+1) + v (step n))
Same for z...
BTW, for the magnus force I supposed a rotation of the ball of 1800 rpm (like a fastball). The results I obtain are similar to the ones given in the sim (93 mph and 35° give me 328 feet instead of 330; 98 mph and 20° give me 323 feet instead of 312; 75 mph and 10° give me 133 feet instead of 141).
By the way, the magnus force has a big impact on the distance travelled: if one takes a rotation equal to zero with a speed of 75 mph and an angle of 35°, the ball will travel
Results
Supposing that the IF needs 1 sec to catch and rethrow the ball, here are some estimations:
ArmeSpeed (mph; OF and IF)___ BreakevenDistance (feet)__MaxDistance (feet)
65__187__193
70__210__216
75__230__240
80__245__262
85__275__286
90__295__312
Distance between C and OF shorter than breakeven => direct throw fastest.
Conclusion
Unless and OF is close to his maximal throwing distance, a direct throw seems faster, based on this model. However some elements here haven't been taken into account such as lateral spin, accuracy of the throw, weather conditions, and all the throws are supposed optimized. Rebounds haven't been considered as well.
For this calculations I used an Excel spreadsheet. If anyone is interested, wants to know how far he can throw, send me a mail or leave a message on the thread. The spreadsheet can also be easily updated in order to include curve, slider, sinkers and screwball.
Sylvain
Posted 9:26 a.m.,
June 17, 2003
(#8) -
tangotiger(e-mail)
Sylvain,
Send me the file, and I'll ask Primer to post it.
Posted 1:46 p.m.,
June 17, 2003
(#9) -
tangotiger
(homepage)
Sylvain's file has been posted to the above link. It is a Excel file. If you have trouble opening it, right-click the "homepage" link, and "save target as".
Posted 12:05 p.m.,
June 18, 2003
(#10) -
tangotiger
Sylvain has issued an update (with graphs!). You can find it at the same link from post #9.
Posted 3:04 a.m.,
June 19, 2003
(#11) -
Sylvain(e-mail)
Thanks a lot for posting it Tango!
Sylvain
Posted 6:08 a.m.,
June 23, 2003
(#12) -
unc84steve(e-mail)
Hi there, I don't post here that often, but I hear good things about this place. I was reading through this thread, and I think the 35 degree angle is way too high because the focus is on distance. Obviously the method is good so far however.
I was going to ask some "real world" questions starting with our maligned LF, Luis Gonzalez (max 10 assists years ago). A few plays come to mind (single with runner on 2nd, digging double out of corner), but I'm guessing the easiest one would be the sac fly, runner tagging 90 feet.
I'd say 4 seconds is a fair 3rd to home time to give. Some batters get flying times of 3.5 down to first, but I don't think runners really want to crash into the "tools of ignorance."
I think 300 feet is a good approximation for a sac fly that runners would gamble on.
My questions center on time & the "lolly pop" throws. MAXIMUM DISTANCE is not really the issue--Gonzo doesn't need to prove to the fans he can throw a ball 300 feet even with a 35-angle. He needs to get the ball there at least within 4.5 seconds so if the runner stumbles, the ball arrives in the same picture frame as the scoring run.
Posted 9:08 a.m.,
June 23, 2003
(#13) -
Sylvain(e-mail)
Steve (I guess): the max distance a throw can reach (given an initial speed Vo depending on the player) seems to be 35 degrees. Ok, I didn't purely demonstrated it... But this has nothing to do with runners being thrown out. The best application of the 35 degrees is hitting.
As for the fielding applications you are right, the relevant issue is time. For your calculations: supposing that Gonzo can throw at 300 feet with an angle of 35 degrees, that makes a speed of about 87-88 mph and 4.6 seconds. And the max distance Gonzo can throw at within 4.5 seconds is about 295 feet.
However a relay with a cutoff man (arm speed also 87.5 mph situated between Gonzo and the Catcher) would take 2*1.36 seconds plus time for the IF to catch and throw (est. 1 sec) = 3.7 seconds. A gain of almost 1 second.
The relevant issues are: given a flyball (as a distance to the catcher) and a runner (time to reach home):
- can Gonzo throw directly to the catcher? If yes, how much time does it take: will he throw the runner out?
- is a relay faster?
The second issue becomes important as soon as the time for the runner to reach home and the time for the direct throw to reach the C become close. However in "live action" it's very difficult to assess all those parameters: is the cutoff man in position, will the throw be accurate enough in order not to lose time, windy conditions. There are a lot of other parameters.
As for your question on the 300 feet as a good sac distance: it depends on the fielders and the runner: as pointed ou above, a relay needs about 3.7 seconds, but a direct throw 4.5 seconds. If one considers a time of 4 seconds for the runner, you have the answer: will the OF go for the relay or not?
Sylvain
Posted 6:25 p.m.,
June 23, 2003
(#14) -
unc84steve(e-mail)
Good guess, it's Steve. It's also coincidental, because I was discussing with a friend how rarely one sees a relay throw on a sacrifice fly play. I really want to see if the first-order assumptions can be made more practical in this model.
Maybe my parameters a little off. Perhaps 300 feet is longer than most sacrifice throw challenges to home plate. Another consideration is that most throws to home arrive on one bounce.
In other words, the real question isn't whether a relay throw is a viable option. It's whether this works on plays we see.
My recollection from college physics is that the time of the play can be solved by looking at the vertical component alone. For example for 250 feet, 4 seconds, assuming ground to ground trajectory, the max height would be the same as the distance any ball would fall in 2 seconds. The key, in one sense, is to keep the throw's max height to a minimum (assuming you can still generate enough of a horizontal velocity/distance). This is one reason that on Astroturf infields, some 3rd basemen developed one-hop throws to keep the maximum height lower (and thus the throw's time lower).
For now, I think it would be good to just test a 250 throw on the fly in 4 seconds to home plate. That would be an impressive Vlad Guerrero or Richard Hidalgo throw. We have a very good idea of pitching mph & angles. You mentioned 35 degrees as one for batted balls. It would be interesting to see if the angles and mph of the throw is reasonable.
Posted 7:20 a.m.,
June 24, 2003
(#15) -
Sylvain(e-mail)
Steve,
Concerning the bounce, well, I don't really know how to look at it. I'd say the loss of energy during the ball-ground collision depends on:
1) speed of the ball (norme)
2) direction of the ball/ground (angle + if grass field, then how the grass grows)
3) nature of the ground: turf/wet grass/sand/dry grass... One could suppose, as you mentioned, that turf might better restitute the speed (harder, different friction coefficient, etc.)
BTW, if the relays aren't used that often, it's because, based on the numbers given in post#7, unless the fielder is really close to his maximal range (nothing to do with RF), a relay would actually cost time. A bounce might even increase this advantage. So the relay is (seems to be barely a viable option). Another point to consider is that when the OF throws the ball, he is running, adding an horizontal initial velocity. I haven't considered this point is my calculations.
Concerning the 250 feet throw in 4 seconds: 82 mph 35 degrees give 3.7 seconds, so that a throw with less speed will reach 250 feet after a rebound and more time, a throw with more speed in less time (85 mph, 250 feet = 3.2 seconds, 28 degrees).
Concerning the vertical component and the may height, I'd say this is correct when neglecting drag, since drag depends on the speed as a whole. But when looking at a throw, it is advantageous to throw with the lowest angle possible: the speed increase due to gravity (also limited by the drag) isn't enought to compensate the difference in distance.
I don't know if I answered your questions.
Sylvain