The wind would slow the ball down and gravity would grab at it too.
If you could make the fastball rise around home plate, you would strike out every batter in every game.
Wrong again, PiC.
What CADad & I are saying can be verified by applying what you will find in any applied physics or mechanical engineering book. If you can't understand those, then look at "The Physics of Baseball" by Robert K. Adair. I quoted from this book, earlier in this thread. But perhaps you didn't read that. Adair is the Sterling Professor of Physics at Yale, and a member of the National Academy of Sciences.
You apparently do not understand fully what you are quoting, and are not able to properly and fully apply the concepts at which you grasp.
What CADad & I are saying can be verified by applying what you will find in any applied physics or mechanical engineering book. If you can't understand those, then look at "The Physics of Baseball" by Robert K. Adair. I quoted from this book, earlier in this thread. But perhaps you didn't read that. Adair is the Sterling Professor of Physics at Yale, and a member of the National Academy of Sciences.
You apparently do not understand fully what you are quoting, and are not able to properly and fully apply the concepts at which you grasp.
My son loved this thread. I learned a lot on this thread.
I asked my son why he has such a good high fb. He responded that he has long tossed for 4 years now and that every since learning physics in hs he has concentrated on getting as much backspin as possible when he long tosses. He says that this has helped him to get extension as well as not having his ball drop like his 2 seamer. His coach calls it "hop", or "late life". He doesn't want to explain the science to a bunch of coaches. lol
There is still a lot to learn.
I asked my son why he has such a good high fb. He responded that he has long tossed for 4 years now and that every since learning physics in hs he has concentrated on getting as much backspin as possible when he long tosses. He says that this has helped him to get extension as well as not having his ball drop like his 2 seamer. His coach calls it "hop", or "late life". He doesn't want to explain the science to a bunch of coaches. lol
There is still a lot to learn.
SBK,
The lift is proportional to the airspeed so you'll get more lift throwing it into the wind. As bbscout noted it'll also slow down quicker and have more time to drop. Kind of like hitting a golf ball into the wind with a lot of backspin, seeing it "balloon" up and then drop straight down. In the case of a baseball I'd guess that you'd actually get less drop throwing across the diamond into a strong wind as long as you were able to throw it pretty hard in the first place. Let's say you were able to throw the ball at 87 mph and the speed dropped by 10 mph on the way to first with no wind. Let's also say you were about 120 ft from first. That means the ball has about 1 sec to drop. Gravity accelerates the ball downward at 32 ft/s^2 and we'll assume you got a lot of backspin on the ball and you've put 12 ft/s^2 worth of lift on the ball. That gives you a net acceleration of 20 ft/s^2 down. That means the ball will drop 10 ft on it's way to first base.
Now lets take the 30 mph wind case. The lift goes from 12 ft/s^2 to about 16 ft/s^2 so the net acceleration is 16 ft/s^2 down. However the ball now slows down by about 16 mph instead of 10 mph as it crosses the diamond (I've made some assumptions here as to the higher drag coefficient with the higher airspeed and haven't really done all the math.) That gives us an average speed across the diamond of about 79 mph instead of 82 mph so the time to get across the field is increased by a factor of 82/79. If we plug that into the distance equation and use a downward acceleration of 16 ft/sec^2 we get a drop of about 8.5 ft into the wind. So the ball would drop a little less but the runner would be more likely to be safe.
What does it all mean? I think that Bighit's son has the right "spin" on this discussion and it is time to stop with the explanations and just throw the #$%^ ball.
P.S. I do realize that the ball would most likely slow down by about 16 or more mph going across the diamond without the wind. The ball would drop a little more in each case but there would still be about a 1.5 ft difference.
The lift is proportional to the airspeed so you'll get more lift throwing it into the wind. As bbscout noted it'll also slow down quicker and have more time to drop. Kind of like hitting a golf ball into the wind with a lot of backspin, seeing it "balloon" up and then drop straight down. In the case of a baseball I'd guess that you'd actually get less drop throwing across the diamond into a strong wind as long as you were able to throw it pretty hard in the first place. Let's say you were able to throw the ball at 87 mph and the speed dropped by 10 mph on the way to first with no wind. Let's also say you were about 120 ft from first. That means the ball has about 1 sec to drop. Gravity accelerates the ball downward at 32 ft/s^2 and we'll assume you got a lot of backspin on the ball and you've put 12 ft/s^2 worth of lift on the ball. That gives you a net acceleration of 20 ft/s^2 down. That means the ball will drop 10 ft on it's way to first base.
Now lets take the 30 mph wind case. The lift goes from 12 ft/s^2 to about 16 ft/s^2 so the net acceleration is 16 ft/s^2 down. However the ball now slows down by about 16 mph instead of 10 mph as it crosses the diamond (I've made some assumptions here as to the higher drag coefficient with the higher airspeed and haven't really done all the math.) That gives us an average speed across the diamond of about 79 mph instead of 82 mph so the time to get across the field is increased by a factor of 82/79. If we plug that into the distance equation and use a downward acceleration of 16 ft/sec^2 we get a drop of about 8.5 ft into the wind. So the ball would drop a little less but the runner would be more likely to be safe.
What does it all mean? I think that Bighit's son has the right "spin" on this discussion and it is time to stop with the explanations and just throw the #$%^ ball.
P.S. I do realize that the ball would most likely slow down by about 16 or more mph going across the diamond without the wind. The ball would drop a little more in each case but there would still be about a 1.5 ft difference.
Just how fast would a baseball have to be thrown to generate enough lift to actually rise?
Just curious.
Lil ole Cessna 172 "lifts" off the ground about 60 knots. Certainly a more effecient "wing" but just not that much speed.
Just a ballpark guess is OK.
Just curious.
Lil ole Cessna 172 "lifts" off the ground about 60 knots. Certainly a more effecient "wing" but just not that much speed.
Just a ballpark guess is OK.
FO, I thought a "knot" was something you got on your arm when a pitcher hit you with a sinker.
quote:
Originally posted by CADad:
What does it all mean? I think that Bighit's son has the right "spin" on this discussion and it is time to stop with the explanations and just throw the #$%^ ball.
That's something I can agree with; I think it was time to do that before this discussion started!
FO,
About 180 mph or twice the rpm or some combination of the two. It is difficult to say for sure because the drag coefficient levels out as you get over 100 mph and the drag goes way up.
About 180 mph or twice the rpm or some combination of the two. It is difficult to say for sure because the drag coefficient levels out as you get over 100 mph and the drag goes way up.
CADad and Texan,
I gotta tell you, I really enjoy the scientific facts you guys bring to the table. Now if I was only smart enough to understand what you're talking about.
I agree that all thrown baseballs lose speed the farther they travel. However I think it's the angle the ball must travel from release point to the strike zone (down hill) that prevents any ball from actually rising.
Be it just appearance or fact, submarine pitchers seem to be able to get more rise... or less drop. I've seen one (submariner) who throws a slider that actually "appears" to go up maybe just a little! He throws low 80s FBs, so it must be the angle more than the velocity.
Being a big believer in scientific facts and knowing very little about them, I'm going with you guys because I'm pretty sure you know what your talking about.
I gotta tell you, I really enjoy the scientific facts you guys bring to the table. Now if I was only smart enough to understand what you're talking about.
I agree that all thrown baseballs lose speed the farther they travel. However I think it's the angle the ball must travel from release point to the strike zone (down hill) that prevents any ball from actually rising.
Be it just appearance or fact, submarine pitchers seem to be able to get more rise... or less drop. I've seen one (submariner) who throws a slider that actually "appears" to go up maybe just a little! He throws low 80s FBs, so it must be the angle more than the velocity.
Being a big believer in scientific facts and knowing very little about them, I'm going with you guys because I'm pretty sure you know what your talking about.
PG,
Submarine pitchers can actually get a little bit of "rise" near the plate. The trajectory is upwards and the drop due to gravity doesn't necessarily completely overcome the upward trajectory and the lift due to spin. That's a special case and one of the reasons submarine pitchers can be so hard to hit even though they don't usually throw as hard.
Submarine pitchers can actually get a little bit of "rise" near the plate. The trajectory is upwards and the drop due to gravity doesn't necessarily completely overcome the upward trajectory and the lift due to spin. That's a special case and one of the reasons submarine pitchers can be so hard to hit even though they don't usually throw as hard.
CaDad,
Thanks for info. I’m still confused because reality doesn’t seem to jive with the math.
You calculated that an 87 mph ball thrown 120 feet would drop 10 feet. When I, or others with good arms throw a ball from short to first, we throw the ball on a straight line.
Thanks for info. I’m still confused because reality doesn’t seem to jive with the math.
You calculated that an 87 mph ball thrown 120 feet would drop 10 feet. When I, or others with good arms throw a ball from short to first, we throw the ball on a straight line.
No one can throw the ball on a straight line. Your perception does not match reality.
I haven't verified CADad's figure of 10', but the ball will drop as it travels 120'.
I haven't verified CADad's figure of 10', but the ball will drop as it travels 120'.
..ain't it amazin' what you can learn on here just observin...............................
Texan, agreed.
It may look flat, but in reality it is thrown on an arc. Even if it is a very small arc, there is a slight uphill release. Just remember that if you throw on an upphill arc, there will be an appearance of carry when using good backspin. It doesn't start falling until the ball reaches the apex of its arc. Not only is their lift, but an uphill force applied and it changes the result.
It may look flat, but in reality it is thrown on an arc. Even if it is a very small arc, there is a slight uphill release. Just remember that if you throw on an upphill arc, there will be an appearance of carry when using good backspin. It doesn't start falling until the ball reaches the apex of its arc. Not only is their lift, but an uphill force applied and it changes the result.
What top-level short stop that you know throws the ball from short to first with a 10 foot drop?
A ten foot drop would be quite a rainbow.
A ten foot drop would be quite a rainbow.
From Adair's book:
"Similar misunderstandings occur about the flight of a throw from third to first by a strong-armed third baseman. Some players emphatically state that a hard-thrown ball travels in a straight line, but a physicist concludes that the ball must drop about 10 feet. Again, the 10 feet is the drop below the initial line of projection. The peak trajectory of the ball is only about 2-1/2 feet above a straight line drawn from the third baseman's hand to the first baseman's mitt. And a 2-1/2-foot deviation over the 127-foot distance between third and first is not far from a straight line."
The key phrase and concept to grasp here is "the initial line of projection."
When you throw across the diamond, you give the ball an initial trajectory ("initial line of projection") that is angled up somewhat. If you released the ball with a perfectly horizontal initial projection, you would one hop the throw.
Hope this clears things up.
"Similar misunderstandings occur about the flight of a throw from third to first by a strong-armed third baseman. Some players emphatically state that a hard-thrown ball travels in a straight line, but a physicist concludes that the ball must drop about 10 feet. Again, the 10 feet is the drop below the initial line of projection. The peak trajectory of the ball is only about 2-1/2 feet above a straight line drawn from the third baseman's hand to the first baseman's mitt. And a 2-1/2-foot deviation over the 127-foot distance between third and first is not far from a straight line."
The key phrase and concept to grasp here is "the initial line of projection."
When you throw across the diamond, you give the ball an initial trajectory ("initial line of projection") that is angled up somewhat. If you released the ball with a perfectly horizontal initial projection, you would one hop the throw.
Hope this clears things up.
SBK,
We already know that the ball drops on it's way to the plate. How much it drops it a function of how long it takes to get to the plate. 120' is twice as far as 60' so the ball takes more than twice as long to get to first base as is does to go from the mound to the plate. The formula for determining how far a ball drops is s = 1/2 * a * t * t, where a is the acceleration and t is the time. Let's say the ball takes 1/2 second to reach the plate and the downward acceleration after you subtract the lift is 20 ft/s. Then you have 1/2 * 20 * 1/2 * 1/2. That equals 2.5 ft. Now make that 1 second instead of 1/2 second since we're throwing the ball about twice as far. It becomes 1/2 * 20 * 1 * 1. That equals 10 ft.
Now, just like Bighit said if you throw the ball upwards just a bit then it will go up 5 feet and drop 5 feet and get to the 1st baseman at the same height that it was released at. If you throw it upwards a bit less, let's say 3 ft upward then it will go up three feet and drop 7 feet resulting in a low throw but not in the dirt. (See next post for correct interpretation)
Finally, yes I have seen a top shortstop make a long throw with an arc on it.
We already know that the ball drops on it's way to the plate. How much it drops it a function of how long it takes to get to the plate. 120' is twice as far as 60' so the ball takes more than twice as long to get to first base as is does to go from the mound to the plate. The formula for determining how far a ball drops is s = 1/2 * a * t * t, where a is the acceleration and t is the time. Let's say the ball takes 1/2 second to reach the plate and the downward acceleration after you subtract the lift is 20 ft/s. Then you have 1/2 * 20 * 1/2 * 1/2. That equals 2.5 ft. Now make that 1 second instead of 1/2 second since we're throwing the ball about twice as far. It becomes 1/2 * 20 * 1 * 1. That equals 10 ft.
Now, just like Bighit said if you throw the ball upwards just a bit then it will go up 5 feet and drop 5 feet and get to the 1st baseman at the same height that it was released at. If you throw it upwards a bit less, let's say 3 ft upward then it will go up three feet and drop 7 feet resulting in a low throw but not in the dirt. (See next post for correct interpretation)
Finally, yes I have seen a top shortstop make a long throw with an arc on it.
What Adair and Texan were saying is that if you threw a ball upwards at 10 ft/s at the same time as you were throwing it at an average speed of 120 ft/s toward first base and there was no downwards acceleration then over the 1 second of travel time to first base the ball would rise 10 ft. and that 10 foot rise is what our drop is relative to. However, there is a downward acceleration of 20 ft/s/s. As a result the ball stops going up after half a second. In half a second the ball has gone from going up at 10 ft/s to not going up so the average speed that it was going up was 5 ft/s. At 5 ft/s for 1/2 second it actually goes up 2.5 feet above the release point. It has traveled about halfway to first base. Now let's apply that distance equation to see how far the ball drops in the remaining half second on it's way to first base. We have s = 1/2 * 20 * 1/2 * 1/2 equals 2.5 ft so the ball drops back down as low as the release point.
We could take this a step further and apply it to the case of a pitcher with a long stride whose stride knee touches the ground such as Seaver used to do. In that case his release point is quite low and he ends up throwing the ball very slightly upward to throw a high strike. The result is a ball that is dropping less as it crosses the plate and a more effective high fastball.
This also explains why a pitcher can throw a more effective 12 to 6 curve with a short stride and a high release point and also why a high curve tends to "hang". A well thrown 12-6 curve released level at a height of 6' will tend to drop about 6'. As it reaches the plate it is dropping at a speed of about 24 ft/s. A 12-6 curve with the same amount of spin on it that crosses the strike zone 3' above the plate had to be thrown up at an initial velocity of 6 ft/s. It will reach it's peak just an 1/8 of a second later just 4.5 inches above the release point for a maximum height difference of 3 feet 4.5 inches as opposed to 6' and it will be moving downward at 18 feet/s as opposed to 24 ft/s as it crosses the plate.
Texan,
Thanks for correcting the error in my previous post even before I posted it.
We could take this a step further and apply it to the case of a pitcher with a long stride whose stride knee touches the ground such as Seaver used to do. In that case his release point is quite low and he ends up throwing the ball very slightly upward to throw a high strike. The result is a ball that is dropping less as it crosses the plate and a more effective high fastball.
This also explains why a pitcher can throw a more effective 12 to 6 curve with a short stride and a high release point and also why a high curve tends to "hang". A well thrown 12-6 curve released level at a height of 6' will tend to drop about 6'. As it reaches the plate it is dropping at a speed of about 24 ft/s. A 12-6 curve with the same amount of spin on it that crosses the strike zone 3' above the plate had to be thrown up at an initial velocity of 6 ft/s. It will reach it's peak just an 1/8 of a second later just 4.5 inches above the release point for a maximum height difference of 3 feet 4.5 inches as opposed to 6' and it will be moving downward at 18 feet/s as opposed to 24 ft/s as it crosses the plate.
Texan,
Thanks for correcting the error in my previous post even before I posted it.
I'm no engineer so don't let me stop you 
, but before I get too bad a headache 
, I am leaning toward
, but before I get too bad a headache , I am leaning towardquote:
just throw the #$%^ ball.
Frank F.,
Sorry for thinking in type. Texan's note explained some things to me that I should have realized but didn't and they've helped me understand some things about throwing a ball that weren't clear to me before.
Sorry for thinking in type. Texan's note explained some things to me that I should have realized but didn't and they've helped me understand some things about throwing a ball that weren't clear to me before.
quote:
Originally posted by Texan:
Wrong again, PiC.
What CADad & I are saying can be verified by applying what you will find in any applied physics or mechanical engineering book. If you can't understand those, then look at "The Physics of Baseball" by Robert K. Adair. I quoted from this book, earlier in this thread. But perhaps you didn't read that. Adair is the Sterling Professor of Physics at Yale, and a member of the National Academy of Sciences.
You apparently do not understand fully what you are quoting, and are not able to properly and fully apply the concepts at which you grasp.
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Even though gravitational force does effect the path of a baseball there is no doubt that the lift of the ball is shown in the forces that are summed in the topspin plots which show the effects of the Bernoulli Principle.
That was my argument and is supported by:
http://math.gmu.edu/~ctombras/projects/proj02/proj02.html
The math involved makes sense, but I must say I've seen a lot of balls thrown 120' without a 10 foot arc. This is based on the on field view, where a 10' arc would be most visable.
For example I've seen many throws from catchers that went right through the vicinity of the pitchers head (or where his head would have been) and reach 2B in the air. When this happens the pitcher is not standing tall on the top of the mound. This distance is actually more than 120'. Please explain the 10' arc on this. I'm also fairly certain that most strong arm 3B and SS's don't have a 10' drop on their throws to 1B. Ozzie Smith is one that did though. Even 3' up and 7' down is not what I'm seeing, almost certain of that.
Still enjoy this discussion a lot, just not sure I understand! I know thrown balls do slow down and drop, I just think there's other factors involved no matter what the trajectory might be.
Also, what about hit balls and the math involved? Can they rise?
For example I've seen many throws from catchers that went right through the vicinity of the pitchers head (or where his head would have been) and reach 2B in the air. When this happens the pitcher is not standing tall on the top of the mound. This distance is actually more than 120'. Please explain the 10' arc on this. I'm also fairly certain that most strong arm 3B and SS's don't have a 10' drop on their throws to 1B. Ozzie Smith is one that did though. Even 3' up and 7' down is not what I'm seeing, almost certain of that.
Still enjoy this discussion a lot, just not sure I understand! I know thrown balls do slow down and drop, I just think there's other factors involved no matter what the trajectory might be.
Also, what about hit balls and the math involved? Can they rise?
Just read Texan's post and that seems more realistic (the 2.5' arc). I think it should be stated that all C, 3B, SS, OFs arms are not the same. Just as some pitchers get more rotation and/or velocity on the ball so do some position players. In scouting terms this is described most often as "carry".
quote:
Originally posted by PGStaff:
Just read Texan's post and that seems more realistic (the 2.5' arc). I think it should be stated that all C, 3B, SS, OFs arms are not the same. Just as some pitchers get more rotation and/or velocity on the ball so do some position players. In scouting terms this is described most often as "carry".
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I would think that "carry' would be optimized using the backspin plots as the model most representative of your question.
In terms of achieving the most backspin rotation and accuracy in flight would be the four seam grip with the fingers located across the large smile and thumb located opposite on the lower distal seam. The ball in flight would have a spin opposite the arm torque created by pronation, which would offset the usual "screwball" effect that one gets when throwing the normal four-seamer.
Like PG, I’m no scientist or advanced mathematician, but I have seen a lot of thrown baseballs and do not believe a ball thrown from short stop to first has a 10 foot drop at 87 mph when thrown over the top from a good player.
I am not saying that CaDad, Adair and others are not accurate in their formula. In fact I’m sure CaDad, Texan and others know their stuff. However when I look at the formula provided by CaDad (s = 1/2 * a * t * t ), I do not see where the different spins of the ball has been factored in.
Pic provided some data on the spin that seems to show how important it really is. PG mentions as an experienced observer how some players seem to have much more “carry” on their ball. I am familiar with shooting a rifle and realize how important the rifling affect has on ballistics.
So when we combine the math provided by our resident numbers expert and combine it with our baseball observation, it appears that a throwers ability to generate backspin make quite a difference. It stands to reason that it makes even a greater difference on longer throws or/and throws being made into the wind.
Perhaps as coaches and parents, we should spend even more time emphasizing better throwing technique that maximizes the backspin and the 6 to 12 rotation. A good drill would be to have players throw directly into a strong wind as they work to maximize the “carry”.
I know I’m going to emphasize it even more and I appreciate everyone’s contribution to this thread.
I am not saying that CaDad, Adair and others are not accurate in their formula. In fact I’m sure CaDad, Texan and others know their stuff. However when I look at the formula provided by CaDad (s = 1/2 * a * t * t ), I do not see where the different spins of the ball has been factored in.
Pic provided some data on the spin that seems to show how important it really is. PG mentions as an experienced observer how some players seem to have much more “carry” on their ball. I am familiar with shooting a rifle and realize how important the rifling affect has on ballistics.
So when we combine the math provided by our resident numbers expert and combine it with our baseball observation, it appears that a throwers ability to generate backspin make quite a difference. It stands to reason that it makes even a greater difference on longer throws or/and throws being made into the wind.
Perhaps as coaches and parents, we should spend even more time emphasizing better throwing technique that maximizes the backspin and the 6 to 12 rotation. A good drill would be to have players throw directly into a strong wind as they work to maximize the “carry”.
I know I’m going to emphasize it even more and I appreciate everyone’s contribution to this thread.
I hate that I missed this topic. Baseball + physics= a great conversation!
I agree that a baseball thrown on a straight line will not rise. But, there may be isolated cases where other factors are involved. Strong atmospheric updrafts or vortices can add additional upward forces to the ball, which may allow them to rise. An example of this is hail. However, if you are playing baseball in conditions such as these, then you will have more to worry about than just seeing the ball rise!
Great topic! Texan and CADad are correct.
I agree that a baseball thrown on a straight line will not rise. But, there may be isolated cases where other factors are involved. Strong atmospheric updrafts or vortices can add additional upward forces to the ball, which may allow them to rise. An example of this is hail. However, if you are playing baseball in conditions such as these, then you will have more to worry about than just seeing the ball rise!
Great topic! Texan and CADad are correct.