Velocity loss of a pitched baseball (June 10, 2003)
Ralph Carmichael works the numbers, and others chime in some corrections. Anyone who wants to do more grunt work, feel free to add it here.
--posted by TangoTiger at 01:09 PM EDT
Posted 11:21 a.m.,
June 11, 2003
(#1) -
Sylvain(e-mail)
If someones here enjoys physics:
http://farside.ph.utexas.edu/teaching/329/lectures/node58.html
It is worth looking at: it explains the curve, screwball, knuckleball.
SC
Posted 12:05 p.m.,
June 12, 2003
(#2) -
JohnW
Robert Adair in The Physics of Baseball addresses this very question. I don't have the book in front of me, but his calculation (I believe) is more precise than the one above.
One problem I noticed (there may be others) with the calculation is that the time of flight of the ball on its way to the plate assumes a constant velocity, i.e. time = distance/velocity. But the velocity is not constant, indeed it's the change in velocity that is being calculated! Also, the drag on the ball is proportional to velocity squared, so as the ball slows down the drag is reduced and the deceleration is also reduced.
Posted 3:55 a.m.,
June 13, 2003
(#3) -
JohnW
Just a follow-up to my previous post. I checked Adair's book and he says that a pitched fastball loses about 7 mph per foot of distance traveled. The effective distance to home plate is somewhat less than 60'6", so the average velocity loss is about 8 mph for a mid-90's fastball. That's actually pretty close to what Carmichael found above, although I think the agreement is somewhat fortuitous.
Posted 6:13 a.m.,
June 13, 2003
(#4) -
Sylvain(e-mail)
There are other limits to the model:
- it doesn't take the gravity into account, i.e. the ball is supposed to follow an horizontal course; the gravity will increase the ball's speed.
- the drag coefficient is not constant. A graph in the link I posted (I pretty much guess it comes from Adair's book, but I don't have the book).
Here is a small improvement of the model presented in Tango's link (so my grunt work):
- we neglect the spin of the ball force; the trajectory of the ball can therefore be described by two variables, x and z, x horizontal, z vertical (no side variations)(with x increasing when going from the pitcher to the plate, z increasing when going up)
- As for the horizontal course, one can suppose that when the pitcher releases the ball (instant t=0), the original speed Vo is purely horitontal (Vox = Vo, Voz = 0)(sorry for the formatting).
Neglecting the vertical drag and the spin, we have: Vz(t) = -g*t (g gravity)
Since the ball needs about 0.4 sec to reach the plate, it gives a maximum vertical speed of 0,5*9,8 # 5 m/s or 11.2 mph. Since there is drag, the actual vertical component of the speed of the ball at the plate will be lower. On the other hand, the horizontal speed will be between 80 and 90 mph, considering the drag or between 42.5 and 35.7 m/s. Vz(plate) is not negligeable, but "small" compared to Vx(plate) (I will consider Vz/Vx <= 1/4)
And the drag can be written: Fd = -a*v^2, with a direction opposed to the speed. Its axial component will be (considering a constant drag coefficient):
Fx = -a*Vx*square_root(Vz^2+Vx^2)
Fz = -a*Vz*square_root(Vx^2+Vz^2)
There fore Fz/Fx <=1/4; considering that at the release of the ball, the vertical drag is 0 (Vo horizontal), the vertical drag is small.
And Fx = -a*Vx^2*square_root(1+Vz^2/Vx^2) can be approximated by
Fx= -a*Vx^2; on a side note Fz = -a*Vz^2*square_root(1+Vx^2/Vz^2)can be approximated by Fz = - a * Vx * Vz
So considering an almost horizontal problem we can write
m*dVx/dt=-a*Vx^2 (1) (and mdVz/dt = -m*g-a*Vx*Vz)
(1) gives m*(1/Vx-1/Vox)= a * (t-to) = a*t (with to= moment of release = 0 and Vox = Vo)
Or: Vx = 1/(a/m*t+1/Vo) = m/a*[(t+m/(a*Vo))^(-1)] = dX/dt
Integrating it one more time:
X-Xo = m/a * (ln(t+m/(a*Vo))-ln(to+m/(a*Vo))
Or X - Xo = m/a*ln(1+a*Vo*t/m)
Considering Xo=0 (origin of the axes = point of release of the ball):
X = m/a*ln(1+a*Vo*t/m)
and t = m/(a*Vo)*[Exp(a*x/m)-1]
Numerical application:
m=145g=0,145 kg
Distance point of release - plate = 17 m (or 55.75 feet)
a=dragcoefficient*cross section area*1/2*density of the air
a=0.3*pi*0.037^2*1,225=0.0006448
With a release speed of 95 mph or 42.5 m/s,
t (ball reaches plate) = 0.416 sec
Horizontal speed (ball reaches plate) = 39.4 m/s = 88 mph
With a release speed of 90 mph or 40.2 m/s
t (ball reaches plate) = 0.439 sec
Horizontal speed (ball reaches plate) = 37.3 m/s = 83.4 mph
Posted 6:18 a.m.,
June 13, 2003
(#5) -
Sylvain(e-mail)
Oups, some incomplete sentences:
"A graph in the link I posted (I pretty much guess it comes from Adair's book, but I don't have the book)" would tend to show that the drag coefficient can be considered as almost constant.
The formatting is bad, so here are the main formulas, if one wants to play with them:
Vx = 1/(a/m*t+1/Vo)
X = m/a*ln(1+a*Vo*t/m)
t = m/(a*Vo)*[Exp(a*x/m)-1]
Sylvain
Posted 4:48 a.m.,
June 17, 2003
(#6) -
Sylvain(e-mail)
(homepage)
From Jayson Stark's latest column:
"The Yankees, of course, had last been officially no-hit on Sept. 20, 1958, by a knuckleballer (Hoyt Wilhelm). So they clearly set one more record Wednesday -- greatest difference in miles per hour of a final pitch of consecutive no-hitters. The Little Unit, Billy Wagner, threw the final pitch of this one -- at about 147 mph. Wilhelm's was more like 57 mph, give or take 20."
147 mph!!!! That seems a lot to me, especially based on the data of the homepage link. Can someone confirm this? Or is this just a typo?
S
Posted 7:49 a.m.,
June 17, 2003
(#7) -
tangotiger
Must be a typo. I think some tennis player just hit 147 mph over the weekend. The max is around 102 or so.
Posted 10:24 a.m.,
June 17, 2003
(#8) -
Sylvain(e-mail)
Based on the spreadsheet I also used for the relay thread:
Suppositions:
* distance home - point of release = 55 feet
* Vo is purely horizontal
* pitch = fastball, no lateral magnus force, rotation = 1800 rpm
Speeds are in mph, distances in feet, time in seconds
OriginalSpeed_SpeedatthePlate_Timeneeded_VerticalDeviation
106__98.4__0.37__-1.4
104__96.4__0.375__-1.45
102__94.5__0.38__-1.5
100__92.4__0.39__-1.6
98__90.3__0.40__-1.7
96__88.2__0.41__-1.8
94__86.1__0.42__-1.9
92__84.0__0.43__-2.0
90__81.9__0.44__-2.1
88__79.8__0.45__-2.2
86__77.7__0.46__-2.3
84__75.6__0.47__-2.5
A loss of 4 mph for a pitcher represents 0.02 seconds more for the batter. I don't think that is negligible. A swing takes about 0.3 seconds to travel between 90 and 120 degres. Supposing that this speed is constant, 0.02 seconds represents a bit less than 10%. But since w the angular speed is growing, it could represent around 10% more rotation! (supposing that the batter begins to swing at both pitches at the same time). And a higher bat speed, meaning more power!
This is of course a very basic view. But a loss of speed of 4 mph is a lot.
This result can also interpret the change represented by an offspeed pitch compared to a fastball.
Sylvain
Posted 3:00 a.m.,
June 18, 2003
(#9) -
Sylvain(e-mail)
As for: "But since w the angular speed is growing, it could represent around 10% more rotation! (supposing that the batter begins to swing at both pitches at the same time). And a higher bat speed, meaning more power!" ,
I should have said: "it could represent around 10% more rotation! (supposing that the batter begins to swing at both pitches at the same time). And a higher bat speed; however one has to consider the loss of momentum of the ball (less speed) as well."
S
Posted 11:34 a.m.,
July 30, 2003
(#10) -
Sylvain(e-mail)
(homepage)
A very interesting link, follow the homepage
Sylvain