Velocity question -- a case of Physics?

I have read that a pitch loses approx 1mph per 7 feet. It will vary on the type of pitche but that is what I have read about a FB (4 seem)

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Do all pitchers' fastballs lose velocity at the same rate? Is this just a matter of physics?

Yes to both. I hope you don't mind if I take this in a purely scientific way. There is no such thing as a negative acceleration in the x direction (the ball litterally slowing down) unless some external force is enacted on it (it is hit, caught etc...). The only force enacted on it in the x direction is air resistance, which in most cases is the minimal of the two main forces. The most signficant force is gravity, which is causing negative acceleration, yet in the y direction (twoards the ground). Gravity is pretty much always the same on earth (at least at the heights where we pitch), and air resistance, as I said, is neglagible compared to gravity.

All in all, they drop in velocity between release and reaching the plate will be almost uniform with very slight variance due to slight differences in air resistance.

doesnt a fastball with more backspin maintain its velocity better?

One philosophy is about release point. The closer you release the ball to the plate, the more velocity is has when crossing the plate, and the less reaction time the batter has from the time it leaves your hand. If you watch pitchers from Oregon State, their coach preaches long strides and has increased stride over 2 feet in some pitchers.

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Originally posted by CPLZ:
One philosophy is about release point. The closer you release the ball to the plate, the more velocity is has when crossing the plate, and the less reaction time the batter has from the time it leaves your hand. If you watch pitchers from Oregon State, their coach preaches long strides and has increased stride over 2 feet in some pitchers.

This makes sense to my unscientific mind. The ball is accelerated by the arm. From the point where the ball is released from the pitchers hand no more accelerating force is being applied to the ball and it is slowing down (I'm not smart enough to know whether it slows at a constant rate). So it would follow that the closer that release point is to the plate, the less distance over which the ball has to slow down. Again, that is my very unscientific interpretation.

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Originally posted by Junkballer:
doesnt a fastball with more backspin maintain its velocity better?

I *think* that the argument there is that the more backspin creates a lifting force to offset some of the force of gravity, thus slightly less resistance and slightly more velocity is maintained. Once again, I'm not smart enough to know how much of this is correct, but I believe that's the basic argument.

Also, when I say "lifting force" I don't mean that the ball actually rises, only that it doesn't drop as fast. I think it's been proven that a human arm can't generate enough backspin and velocity to actually cause the ball to rise, but again, I'm not positive.

The back spin is not enough to cause a FB to rise. Spin creates resistance and slows the ball in its forward motion. The weight of the ball and the forward resistance will cause the ball to drop.

If you look at the gameday results you'll see that pitches thrown in the low 90s tend to lose between 8 and 9 mph, while pitches thrown in the low 80s tend to lose 6 to 7 mph.

I don't know if this is due to a higher spin rate on breaking pitches or if it is due to the velocity squared term in the drag equation. My guess is a little of both, but that is just a guess.

You also have to realize that the drag coefficient for a spinning baseball goes down with increased velocity, but it is flattening out as you get into velocities in the 90s and above making the V^2 term more important.

BTW, a fastball thrown to the top of the strike zone from a low release point can be still rising slightly as it reaches the plate. The rise is coming from the initial trajectory. The extreme example of this is a submarine pitch which can be still rising despite forward spin as it reaches the plate simply because it was released at such an upward angle. I know many of you have read studies that "proved" that a fastball can't be rising as it crosses the plate, but I have yet to see a study taking into consideration the case of an overhand fastball thrown from a low release point toward the top of the strike zone, ala Tom Seaver. I've done the math and it works, although one does have to assume spin rates on a fastball similar to those of a curve ball for it to work.

It would have been interesting to see the velocity change on Wakefield's knuckler vs. his fastball as the knuckler with so much less spin should slow down a lot more.

As it turns out the gameday results were still up on mlb and Wakefield's mid to high 60s knucklers to the first batter dropped an average of over 7.2 mph while his mid 70s fastballs dropped an average of 6.4 mph. That tells us that the trend is for more spin to result in less velocity drop. On the other hand his fastballs had less of a velocity drop than Byrd's mid 80s fastballs so the velocity also comes into play.

As far as a release point being closer to the plate making a significant difference, it doesn't. As we've seen the velocity drops fairly uniformly along the way to the plate and if we assume a distance of 54' vs 52' as pitchers do release the ball well in front of the rubber although probably not that much, then the velocity difference would be about 8 mph/27 which comes to about 0.3 mph. The difference in reaction time is also not a significant factor for the same reason. In other words, it may help a pitcher a little bit but if it causes even a 1 mph drop in release velocity it isn't worth it.

willj,
JMO, but fastballs that explode on a hitter are more likely to be the result of deception in the motion than differences in how much a ball slows down or release distance. A motion that convinces the hitter that a slower pitch is coming or a motion that makes it difficult to pick up the pitch is the more likely cause.

Even in the case of the submarine fastball, I would imagine that a fastball drops relative to it's initial trajectory, which I think should be the point of reference. Obviously we can take this example to the ultimate extreme and imagine throwing a fastball directly into the air, so that relative to the ground the fastball would certianly rise in that instance. But I think what we're really talking about is the ball rising/dropping relative to the initial trajectory.

The studies I referenced "proved" that a human cannot throw a fastball hard enough and with enough backspin that the "lift force" (I don't know what the scientific term would be) generated by the spin and velocity can make the ball rise. For more technical terms look up the magnus effect, which I understand but wouldn't begin to know how to explain.

Another example would be what softball pitchers call their rise ball. The ball doesn't actally take off and shoot upwards or anything like that. It's just that the pitchers arm motion causes the ball to cross the plate at a higher point than where it left the pitcher's hand (I'm sure pitching off a flat mound helps here). But even in this case I would guess that the ball drops relative to the initial trajectory. No way am I smart enough to figure out if any of this is correct though.

It is a given that a ball drops relative to it's original trajectory. A good curve generates about 16 ft/s^2 acceleration. If we assume that the backspin on a fastball can generate that much acceleration upwards then the ball is still accelerating down at 16 ft/s^2 meaning that in a half second it is going to drop about 2 ft relative to it's initial trajectory. The reality is that few, if any fastballs have that much spin and the ball drops a bit more relative to it's initial trajectory. However, any fastball that ends up high in the zone is thrown with a bit of an initial upwards velocity that has to be overcome.

Submarine pitchers throw the ball in an upward directioncompared to the horizonatal plane. The ball goes up sligthly because the pitch is thrown that way but it drops dranatically and that is what makes their ball hard to hit. It has been proven that FBs do not rise. It also has been proven that a FB slows down approx 1 mph every 7 feet on average to the plate. That is the average over the 60'6" thrown by pro pitchers.

The 1 mph every 7 ft. was done by Adair and was based on analysis. The real numbers can be found by looking at the gameday pitch speeds.

CADad, the spin of the curve is generating a force acting in the same direction as gravity. The spin of the fastball is generating a force that is fighting against gravity.

If by low release, you mean submarine, I can go with you. Low three quarter? No.

And you and EH are correct, any pitch thrown will go down when compared to its original vector at release.

Texan,
By low release I mean a pitcher like Seaver with his knee on or near the ground at release and a low 3/4 delivery. That could be as low as 3 ft. off the ground. Now assume a tall hitter and an umpire with a high strike zone, so 5' above the ground for the top of the strike zone. Now assume a ball going an average of 90 mph and traveling 54' to the front of the strike zone, since the pitcher releases out front of the rubber and the plate is about 17" deep. We can round that to .41 seconds travel time. Let's assume the upward acceleration imparted by the spin is 16 ft/s^2 so the net downward acceleration is 16 ft/s^2. That means the change in velocity due to gravity and spin is 16 ft/s^2 * .41s = 6.56 ft/s.

Now in order for the ball to be still rising after 54' the initial upward velocity of the ball would have to be greater than 6.56 ft/s. So let's assume that it is 6.6 ft/s such that the average upward velocity over the flight of the ball is just over 3.3 ft/s. Over 0.41s the ball will travel upwards 1.35 ft. 3' + 1.35' = 4.35' which could be the top of the strike zone with the ball still moving slightly upwards. Now one can question if the acceleration due to spin can really reach 16 ft/s^2 and if any pitcher actually releases the ball as low as 3' above the ground. On the other hand it is easy to see that an eye high pitch, the kind power pitchers get a lot of K's with could easily be still rising as it crosses the plate.

It is some of those assumptions I am questioning.

Give me the right to make the assumptions, and I can prove anything. 5' is awfully high for the top of the zone, even with a 6'6" hitter.

What is behind the assumption of the spin giving the 16 ft/s^2 acceleration?

And I don't think even Seaver released 3' above the ground.

Etc, etc....

Texan,
A well thrown curve breaks about 18" independent of gravity. It takes an acceleration of around 16 ft/s^2 to produce that on hard curves. It wouldn't surprise me in the least if Pedro Martinez gets that much spin on his 4 seam fastball, although I doubt most pitchers could. However, take a few ft/s^s off that and you aren't going to get a major change in the result.

Go to the release position with your back knee on the ground, remember that when a pitcher is at that point there's very little height left on the mound and put your arm at a low 3/4s release position. You'll be surprised how low that is.

Yep, 5' is high, but you'll notice the ball was at 4.35' and still rising slightly.

I'll agree that my assumptions are optimistic and the probability of it really happening are slim but it isn't impossible and the real meaning of this for a pitcher is that a high fastball thrown at or a bit above the top of the strike zone can react very differently from a low fastball and that's why pitchers can get the strikeouts on a high fastball and can get swings on pitches that are up out of the zone.

CADad,
Well, I tried to see how low I could get, and established that at my age, it is an uncomfortable position! But the best I could do was 4 ft above the ground. I really don't think that even Tom Terrific could launch a 4 seamer from 3 ft above the ground.

The figure of 16 ft/s^2 is twice as high as Adair's estimate of 1/4th of gravity. Furthermore, Briggs used a vertical wind tunnel to show that the drag at 95mph is equal to the force of gravity. It is very hard for me to believe that half of the total drag can be directed perpendicular to the direction of motion.

Most MLB baseball parks have cameras located near 1st and 3rd base. These are well positioned to show the track of a pitch in a vertical plane. I strongly believe that if any pitcher were even close to producing a "rising fastball", we'd be seeing it on broadcasts.

Finally, I'm puzzled by the idea that high fastballs react differently than low fastballs. The aerodynamics should be very similar, and the deviation from the initial trajectory should hardly differ. I think batters swing at high fastballs because they expected or perceived a slower or downward breaking pitch.

Low fastballs are moving at a fairly significant downward angle due to both the drop due to gravity and the initial release angle being about level.

High fastballs were released with an upward velocity and as such have little downward movement as they approach so hitters tend to swing under high fastballs. When pitchers can release from a low release point with a lot of backspin then the amount the ball is dropping is so much less than is normal for hitters that they have trouble dealing with the pitch.

OK, so let's assume Adair was using a typical number and the actual number for a power pitcher getting a lot of backspin is in between 16 ft/s^2 and 8 ft/s^2 = 12 ft/s^2. Let's also assume a 4 ft high release point. Then the change in velocity is 20 ft/s^2 * .41s = 8.2 ft/s. Let's start the ball with an 8.2001 ft/s initial velocity upwards. Then the ball will have an average upwards velocity of 4.1 ft/s upwards for .41s resulting in a 1.68 ft rise and an extremely small upwards velocity when the ball is 5.68 ft above the ground. That is up out of the strike zone for virtually every hitter but still something hitters swing at.

Compare that to a level release from a more typical 5+' off the ground where the ball is going to be dropping at 8.2 ft/s as it reaches the plate and that's assuming a lot of backspin.

Now let's assume an initial upwards velocity of 6 ft/s and a release point 4' off the ground. The ball will be dropping at 2.2 ft/s as it reaches the batter and the average velocity will be 1.9 ft/s upwards for .41s resulting in about a 0.8 ft rise from the original 4 ft. release point. So a fastball at the top of the strike zone from a pitcher with a low release point throwing a 4 seamer can be expected to be dropping at 2.2 ft/s as it approaches the plate while a low fastball can be expected to be dropping at 8.2 ft/s or more. That is a big difference for a hitter to adjust for and the reason why hitters tend to swing under high fastballs if they aren't ready for them.

We could also assume a 100 mph pitch averaging 95 mph which would reduce the time slightly and make it more feasible. My guess is that Nolan Ryan got a lot of backspin on his 4 seamer and his release wasn't as low as Seaver's but it was fairly low.

I should have given a little more detail in mentioning Adair's estimates. He calculated that a baseball spinning at 1800 rpm, which he claims is the highest rate pitchers actually can achieve, would have a Magnus force equivalent to 1/4th the weight of the ball for any pitch speed between 50 and 105 mph. He also states that he believes his estimate is accurate to within 25%. His believes his estimate is bolstered by noting that sidearm pitchers, who generate Magnus force in the horizontal plane, thus separating the effects of aerodynamics and gravity, can only achieve about 8" of horizontal break. That corresponds nicely to 1/4 of gravity.

Is he right? I think a number of MLB parks have the equipment in place to make the measurements routinely, so maybe we'll find out in the next year or so.

3fingered,
Sounds like he's right within limits, but I'd still guess that Pedro and a few others can get more break than that. I don't know how much more so I don't know if they'd fit within his 25%.