# A question on boundary layers viscosity and air seperating frmo a ball

1. Dec 21, 2007

### Omegared

a question on boundary layers viscosity and air seperating from a ball

I have a few questions that have to do with a viscosity on the surface of an object, the boundary layer and the boundary layer separating from the surface of a baseball!

1) my first question is if we had a stationary object lets say a baseball on a table would the baseball have a boundary layer? Or would the baseball have to be thrown for it to have a boundary layer?

2) My second question has to do with the boundary layer separating from the surface of a base ball as seen in the diagrams below (the ball is not spinning in both diagrams)

Here is a more detailed diagram

Now I read that a sphere like a baseball is a poor aerodynamic shape and if it is not spinning the boundary layer will separate similar to the figure above creating a large turbulent wake of low pressure at the back while the front has high pressure so the front to back pressure difference creates a backward force on the ball ie drag.

a) i was just curious could this pressure difference on the front and back of the ball be another application of the Bernoulli principle?

b) there was something in the explanation of the diagram above (figure2) that I just didn’t understand first of all here is how the diagram was explained:
As smooth air flows past a ball, it experiences a uniform increase in static pressure in front of the ball, a decrease in pressure along the sides, and an increase in pressure towards the back. As the air pressure increases toward the back, the air molecules lose their ability to follow the curvature of the ball (as they try to fight the increasing pressure).

The cause of this separation happens because As the flow moves downstream from the 90° or 270° position, it encounters an increasing pressure. Whenever a flow encounters increasing pressure, we say that it experiences an adverse pressure gradient. The change in pressure is called adverse because it causes the airflow to slow down and lose momentum. As the pressure continues to increase, the flow continues to slow down until it reaches a speed of zero. It is at this point that the air no longer has any forward momentum, so it separates from the surface. separated flow creates a region of low pressure in the wake!

i) fist of all why is there a “decrease in pressure along the sides” of the ball?

ii) the above explanation also says As the flow moves downstream from the 90° or 270° position, it encounters an increasing pressure…………As the pressure continues to increase, the flow continues to slow down until it reaches a speed of zero. It is at this point that the air no longer has any forward momentum, so it separates from the surface

Where is this Increase in Pressure along the 90° or 270° Coming From? Is it because the air is trying to fallow the curved surface but cannot because the shape of the ball is unaero dynamic?

3) this is something that really got me a bit frustrated the above explanation in green says that reason the airflow separates is because the “flow continues to slow down until it reaches a speed of zero. It is at this point that the air no longer has any forward momentum so it separates”

And yet I found another huge article (over 25 pages long) that explained the separation of air flow from the ball in a in a completely different way it says “if the air is moving fast enough past the ball the air molecules cannot move back behind the ball in a laminar or stream line flow to fill in the space behind the ball the result is turbulence. This inability of the air molecules to quickly follow behind the ball causes a partial vacuum behind the ball (I take it that by partial vacuum he is referring to the turbulence)

Now believe me I am far far far from an expert in physics but these two explanations regarding why the air separates from the surface of the ball seem to be totally different!!! The explanation in green seems to say that the air separates because it is moving too slow! And on the other hand the explanation in blue seems to be saying that the air separates because the air is moving too fast? Am I misunderstanding something???

I have a few more questions ( including the viscosity part) but so I don’t deter anyone from answering my questions I’ll post them on a different thread a little later! If anyone can help me out could you please put the number of the question your answering near your answer 1, 2a bi, bii, 3. so I don’t get confused thanks

If anyone can help me out please do I really need it and try to keep in mind I am far from an expert in physics

Last edited: Dec 21, 2007
2. Dec 21, 2007

### Astronuc

Staff Emeritus
The boundary layer only makes sense in the context of motion of the fluid with respect to and object or surface, or with respect to motion of an object in a fluid. The first situation would be like air in a wind tunnel flowing past a model aircraft, or fluid in a pipe, and the second situation is that when a ball or aircraft travels through the air. The fluid-object interaction and fluid dynamics is similar.

Also, not only is a spherical shape relatively poor aerodynamically, from the standpoint of drag, by cylinidrical geometries are also poor aerodynamically, and vortex shedding is also an issue, particularly in heat exchanges that employ parallel tubes in cross flows of high Reynolds number.

I'll write more later when I have time, but meanwhile hopefully others will contribute to the discussion.

3. Dec 22, 2007

### caslav.ilic

Boundary layer is not a physicall phenomena as such, but rather a convention used to describe and model flows. Not going into definitions of boundary layer, basically it serves to classify the flow areas into a part that is reasonably dependent on viscosity (inside boundary layer) and the part which behaves mostly as inviscid flow (outside boundary layer).

For a zero-speed object, the definition of boundary layer size (thickness) would not apply. But, the slower/smaller the object is for the given fluid, the bigger its boundary layer is going to be. Thus, for a zero-speed object, you could say, as a limit, that the boundary layer is infinite, i.e. all flow is boundary layer. Albeit the concept of boundary layer becomes of no practical use long before that.

No, as the Bernoulli theorem holds for inviscid fluid.

The Bernoulli theorem is sometimes used for viscous flows precisely when the boundary layers are small and well-behaved with respect to the object, so that the pressure variations can be mostly modelled as inviscid -- such as in some cases of air flow around an airfoil of airplane-worthy size and speed.

The flow around a ball is decidedly unlike inviscid flows, and the assumption of inviscid flow and usage of Bernoulli theorem would give that the pressure difference between the front and the back is zero. Sometimes the Bernoulli equation is empirically "uprated" for the viscosity, such as in pipe flows, but that still requires well-behaved flows, which must be experimentally well-examined.

In my opinion, this question is not exactly correctly posed. Why is the speed of light the upper limit? Why is gravitational force inversely proportional to square of distance? Why is... Because they have just been assumed/observed to be, and theories built upon such assumptions have shown prediction capability.

So with the fluid flow, based on observations, there are equations of rising complexity which have been formed to model it. The simplest of them, potential flow equation (for which Bernoulli holds too), has an analytical solution for the sphere which will show this 90°/270° peek of underpressure. In particular, the realistic, viscous flow about a sphere which you are pondering about, will start by reproducing the potential flow solution up to some angle, and then veer off due to viscous effects. But, the pressure will still grow to some point, then start reducing, and at some point the flow is going to go ugly, "boundary layer separating".

That's just how the flow is, no why's :) The important bit is that people have been able to come up with some generalizations (equations) based on observing such facts.

They are not really different, just the second one is throwing into the mix another observed characteristic of viscous fluid flows -- turbulence. Don't think of turbulence as that which throws you about in an airplane, a pack of rough air or something. In this context, the turbulence is a particular flow feature, which has been observed to happen when the object of given size reaches some speed. This is why the second explanation is having the "fast enough" part.

However, turbulence is not responsible for the boundary layer separation. It will happen whether there is turbulence or not. The rise of turbulence only influences how and when it happens. It goes like this: flow is very slow, separation is very well past the 90°/270° point; it starts to take up speed, separation starts moving forward, and it comes in front of 90°/270° point; sometimes in the meantime (depending on the size of the object), the turbulence may appear, which suddenly pushes the separation point again to the back.

What's this stuff about turbulence then? Well, the equations which model viscous fluid flow, Navier-Stokes, happen to be non-linear. This means they can have many solutions, but mysteriously the nature chooses only some of them -- and there are experiments where people "force" the nature to provide one or the other solution for the one same flow setup. And then a lot of talk how to compute solutions to Navier-Stokes equations such that they correspond to those in nature...

--
Chusslove Illich (Часлав Илић)

4. Dec 23, 2007

### Omegared

thanks for the help

thanks for answering if any one els can contibute please do i have some follow up queestions to some of the responses that i will post here a little later until then if anyone else would like to contribute please do!

5. Jan 2, 2008

### sayak

I will basically say the same thing as above. If you consider inviscid flow only then from 0 to 90 pressure decreases and then from 90 to 180 the pressure again increases in such a way that pressure at 0 (front of ball) and pressure at 180 (back of ball) is same. There are analytical solutions that predict that this to be so. However the picture becomes more complex when you factor in viscosity. From 90 to 180 the fluid is moving through adverse pressure gradient. So pressure force is acting against the flow direction. Coupled with this we have the viscous force that obviously always acts against the flow direction. Now, very near the surface of the ball, velocity is necessarily small (because at the wall, we have u=0 due to no slip condn., where u is the velocity component tangential to the ball surface). Hence here the inertia force(or mommentum) of the fluid is small. So the combined effects of adverse pressure gradient and viscosity is successful in decelerating the fluid, stopping it and reversing the flow direction in the near wall region of flow. This induces a pile up of fluid (since fluid downstream is still moving forward along the ball's surface) and hence seperation of the boundary layer from the ball's surface. The net effect of seperation and wake formation is that the pressure does not recover to the upstream value at the back of the ball (ie at 180). And hence there is a net pressure difference between the front and the back side that leads to AERODYNAMIC DRAG that you may feel while riding a motorcycle. That's the gist. Now where seperation will occur depends on the shape of the body, the fluid and of course the flow itself. This is a huge subject of study because an aerofoil is after all a cleverly designed body made such that this net aerodynamic drag has a vertical component that generates the requisite lift force.

6. Jan 6, 2008

### Omegared

i think i finally got it!

Thank you very much for getting back to me. I have been going over your response and some other websites and though physics is not one of my strong points I think I started to understand what is going on. And I would just like to check with you to see if I understand why the boundary layer separates from the surface of a baseball. Ok here it goes:

1) In thee case of a baseball thrown with no spin (like in the first diagram below) when the airflow reaches about the half way point around the ball the bend becomes too tight for the air flow to follow the curved surface this creates a lot of high pressure at the top and bottom of the ball that the viscosity of the air can’t support so the boundary layer separates. Did I get this right?

2)On the other hand if we were throwing the baseball with a lot of spin (Like in the diagram below) this would change the pressure around the ball. On the top side of the ball we have the ball’s surface spinning in the same direction as the air flow this would speed up the air flow on that side and decrease the pressure so this would allow the ball to pull the boundary layer further around the top side of the ball before separating. On the bottom side of the ball we have the ball’s surface spinning in the opposite direction of the airflow creating a lot of friction and slowing down the air flow, this would increase the pressure on the bottom side of the ball and the boundary layer to separate earlier. This deflects the wake downward and according to Newton's 3rd law the wake being deflected downward is the action and the reaction is a lifting force upward. And there is also a lifting upward due to the pressure differences acording to bernoulli's principle So did I get this right?

thank you very much and i hope to hear from you soon

7. Jan 6, 2008

### sayak

yes you did. that is why a spinning ball swerves in the air.

8. Jan 10, 2008

### Omegared

what happens if we add more speed to a spinning baseball

thanks for helping me out

Ok I understand that a spinning ball creates lift. And that’s what makes a baseball curve But I am kind of curious how to increase the lifting force and create more lift (obviously just add more spin) but I also read some information on Curve Balls that if the ball is thrown faster this will create a little bit more lift for the same amount of spin causing it to curve better.

Speed
ft/sec ......Spin rpm...........Max Curve,in inches from a straight line

75.............1200..................10.8
100...........1200................... 11.7

The effect of speed got me a little curious. I know more lift can be created if more air is diverted faster based on a variation of Newton’s second law Force=mass x acceleration

I took a curve ball curving to the left for an example, would more lift be created because the faster the ball is thrown the faster the air flows over the left side of the ball which is spinning in the same direction as the air flow, this decreases the pressure on that side and allows the ball to pull the boundary layer further around the left. on the other hand on the right side of the ball the ball's surface would still be spinning in the opposite direction creating high pressure causing the boundary layer to separate earlier. So the wake would be deflected a little more to the right. ie more air is deflected faster to the right so we get a greater lifting force to the left based on a variation of Newton’s second law Force=mass x acceleration.

Is that the little extra lift more speed gives to a spinning baseball

Last edited: Jan 10, 2008
9. Jan 10, 2008

### Astronuc

Staff Emeritus
The lift simply has to do with the differential pressure across the ball (between top and bottom). The rotation compresses air under the ball. The compression comes from the forward speed, so increasing the speed compresses the air just a little more. Rotating faster maintain slightly more compressed air under the ball - that is before the air molecules diffuse/flow away from the ball into the lower pressure air sorrounding it.

At the surface of the ball, the air has zero velocity, which is the reason a boundary layer develops. The motion of the ball (or any object) compresses the air with which it interacts, and the air then moves around the object while pressure and velocity gradients develop.

The objective with lift is to get the air underneath to compress while the air on top is either not compressed or slightly reduced in pressure. The differential pressure gives the lift by creating a net vertical upward force.

In the example given the rotation moves the air quickly over the top so that the compression is less, while on the bottom the air is slightly more compressed. The downward wake comes from the air being pulled down at the surface of the ball, and that deflection (acceleration) also adds to the lift.

Are you now talking about a ball spin around a rotational axis parallel to the velocity. That would be different than the one shown in the illustration.

10. Jan 12, 2008

### Omegared

maybe i my question wasn't clear enough (sorry). My problem wasnt with the speed of spin of the ball. But how the actual increase in forward speed of the ball can make a baseball curve better for the same amount of spin as implied in this site right here http://www.100.nist.gov/curverelease.htm [Broken] base on work in the 1950's.

something else that i found a bit confusing was that the site above clearly says that an increase in forward speed of the ball can make a baseball curve better for the same amount of spin in the 60ft between batter and pitcher. And yet other sites seem to imply that this is not possible because even though the lifting forces are increased by throwing the ball faster it also gives the forces less time to act so the ball actually has less of a curve.

So i would like to know if indeed an increase in forward speed of the basebball can make a baseball curve better for the same amount of spin. is my interpretation above correct ie the faster the ball is thrown( ie forward speed) the faster the air flows over the side that is spinnin in the same direction as the airflow so we have reduced pressure on that side allowing the boundary layer to flow further around the left and diverting more air to the right for a greater lifting force left

Last edited by a moderator: May 3, 2017
11. Jan 12, 2008

### Astronuc

Staff Emeritus
Drag increases with forward speed, and the compression of the air would increase with the speed, which would improve the curve of the ball.

12. Jan 13, 2008

### caslav.ilic

Both statements are correct. The ball has mass, so it takes time for the lift force to change its path sideways. Higher initial velocity means higher lift, but less time to the target; and vice versa. For a given ball, amount of spin, and distance to the target, there exists one optimal initial velocity that will give the most curvature to the path. Don't even think of asking how to compute that velocity :)

In practice, given that the lift force grows with the square of velocity, and time to the target drops only linearly, it may be that in the given baseball scenario noone can actually throw so fast to go over the optimal initial velocity, hence the "throw as fast as possible" advice.

--
Chusslove Illich (Часлав Илић)

13. Jan 13, 2008

### Omegared

a lot of sites explain the pressure differences on the top or bottom of a ball as the result of the ball changing the speed of the air. For example a spinning ball like in my blue diagram above would have the top spinning in the same directrion as the airflow speeding it up causing low pressure and at the bottom the ball would be spinning on the opposite direction as the airflow slowing it down causing high pressure.

so when you say the air is "compressed" is it the same as saying slowed down?

so in regaurds to forwaed speed effecting lift would i be correct in saying:

Speed has little to do with improving the lifting force on a ball but it helps a little. The faster the ball is thrown the faster the air flows over the side of the baseball spinning in the same direction as the airflow so there is a little less pressure on that side. And on the other side of the ball we have the faster flowing air flowing over the side spinning in the opposite direction, this cauises a little more friction slowing the air down and a higher pressure. so we get a greater lifting force!

Last edited: Jan 13, 2008