HobbyKing.com New Products Flash Sale
Reply
Thread Tools
Old Nov 15, 2012, 07:32 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
The below examples help me understand the (fairly subtle) difference between a system in which the net momentum is changing and a system in which the net momentum stays the same…

Imagine you are standing next to a long, straight, level stretch of railroad track (frictionless of course) with a string of boxcars on it extending for a mile in either direction. If you were to exert a force of magnitude F on one of the boxcars, and the direction of the force was to the right as you faced the track, then Newton’s Second Law says the string of boxcars would accelerate to your right at a rate of F/m (where m is the total mass of the string of boxcars). You would experience a “reaction” force to your left of magnitude F. Just basic Dynamics.

Now imagine that you took the track directly underneath the string of boxcars and bent it into a circle with a circumference of 2 miles (so you’d be able to connect the first and last boxcars). If you were to again exert a force of magnitude F (to your right) on a boxcar, the result would appear very much the same. The boxcars in front of you would accelerate to the right at a rate of F/m, and you would experience a reaction force to your left of magnitude F. There's a difference though...

In the case of the straight track, the center of mass of the string of boxcars accelerates to your right at a rate of F/m (the string experiences net momentum change). In the case of the circular track, the boxcars in front of you accelerate to your right at a rate of F/m, but their counterparts on the opposite side of the circle accelerate to your left at a rate of F/m. The center of mass of the circular string of boxcars remains at the center of the circle (there’s no net momentum change). The harder you push to the right, the harder the track pushes to the left to keep the train on the track. The circular track illustrates that it’s possible to achieve a reaction force by pushing against an object (and putting that object in motion), but without transferring net momentum to that object. All that is required is that you cause something else push on that object with equal magnitude in the opposite direction.

When you exert a force on the circular string of boxcars, you are changing the momentum of the individual boxcars, but when you sum up the momentum change of all the boxcars that are in motion, the net momentum change is zero (momentum is a vector quantity). In this case it is not accurate to say that the force you exert is accompanied by a change in the momentum of the string of boxcars. Instead, the force you exert is accompanied by a reaction force between the string of boxcars and the railroad track.

If you don’t see a fundamental difference between the above examples, then there’s not much point in trying to connect this analogy to a wing pushing against the air. On the other hand, if you see that the straight track example represents momentum exchange and the circular track represents balanced reaction forces read on…

Which of the above examples is a better analogy for a lifting wing in steady flight? If you look at the transverse flow pattern in the wake of a lifting wing, it’s pretty apparent that the downwash travels not straight down, but in closed (or nearly closed) paths. While this is in no way a quantitative proof that there is or is not net momentum left in the wake, it certainly suggests the possibility that there is more at work here than pure vertical momentum exchange.

A quantitative analysis of the momentum in the wake of a lifting wing can be done, and it shows that there is not always equality between the lift and the air’s rate of momentum change.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Sign up now
to remove ads between posts
Old Nov 15, 2012, 06:15 PM
t=1007606
HX3D014's Avatar
Joined Dec 2006
423 Posts
Quote:
Originally Posted by richard hanson View Post
All very esoteric - but how it relates to model science -simply escapes me.
I looked up the word Science and it is from the Latin word "scientia," meaning Knowledge.

I only imagine that those who participate in the thread are Either trying to impart some knowledge or further Gain some .

Bryce.
HX3D014 is offline Find More Posts by HX3D014
Reply With Quote
Old Nov 15, 2012, 06:29 PM
t=1007606
HX3D014's Avatar
Joined Dec 2006
423 Posts
Shoe, There are different types of Momentum in the conservation of Evergy analogies.
One is translational and one is Angular.
HX3D014 is offline Find More Posts by HX3D014
Reply With Quote
Old Nov 15, 2012, 09:37 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
. and the circular track represents balanced reaction forces read onů.
What more could we possibly want!! lift comes from balanced action/reaction.
Momentum is just book keeping.

Regardless of how one feels lift to be created, centripetal accelleration exists were flow turns. Fluid static pressure is changed as the fluid mass inertia resists (reacts) to that acceleration. Since, on the wing, that acceleration is normal to each surface element the pressure lift contribution is equal to the vertical component of the that mass acceleration. When all is summed up the lift is equal to the vertical comonent of the total mass acceleration in turnig.

What happens aft of the TE is not basic to creating lift.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 15, 2012, 11:33 PM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
Quote:
Originally Posted by HX3D014 View Post
Shoe, There are different types of Momentum in the conservation of Evergy analogies.
One is translational and one is Angular.
When Newton says that an object's rate of momentum change is equal to the unbalanced force acting on it, he is referring to the object's linear (or translational) momentum, not it's angular momentum.

I don't think anyone has yet tried to suggest that the lift force acting on a wing is equal to the air's rate of angular momentum change (or the air's rate of energy change).
ShoeDLG is offline Find More Posts by ShoeDLG
Last edited by ShoeDLG; Nov 15, 2012 at 11:41 PM.
Reply With Quote
Old Nov 15, 2012, 11:54 PM
t=1007606
HX3D014's Avatar
Joined Dec 2006
423 Posts
Quote:
Originally Posted by ShoeDLG View Post
When Newton says that an object's rate of momentum change is equal to the unbalanced force acting on it, he is referring to the object's linear (or translational) momentum, not it's angular momentum.

I don't think anyone has yet tried to suggest that the lift force acting on a wing is equal to the air's rate of angular momentum change (or the air's rate of energy change).
It applies to the Angular.

Now imagine your train track was a free body, and you applied a force in one direction same as before. But not for a full revolution maybe for just 5deg of the rotation of your train link. This time you will see an angular and translational momentum result.
HX3D014 is offline Find More Posts by HX3D014
Reply With Quote
Old Nov 16, 2012, 04:27 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
Quote:
Originally Posted by HX3D014 View Post
It applies to the Angular.

Now imagine your train track was a free body, and you applied a force in one direction same as before. But not for a full revolution maybe for just 5deg of the rotation of your train link. This time you will see an angular and translational momentum result.
I don't understand what you're saying. If you have a string of boxcars that form a closed loop, there is no way for the translational momentum of that string to change (as long as it stays on the track).
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 16, 2012, 06:13 AM
greg
ciurpita's Avatar
somerset, nj
Joined Feb 2005
371 Posts
Quote:
Originally Posted by ShoeDLG View Post
I don't understand what you're saying. If you have a string of boxcars that form a closed loop, there is no way for the translational momentum of that string to change (as long as it stays on the track).
... on what scale?

Quote:
Originally Posted by ShoeDLG View Post
When Newton says that an object's rate of momentum change is equal to the unbalanced force acting on it, he is referring to the object's linear (or translational) momentum, not it's angular momentum.
i've been interpreting "rate of momentum change" as m dV/dT = ma = F. And when you say there is little or no rate of momentum change as being that ma = F = ~0. but certainly if you put something into motion, even if it's just angular motion, a Force is involved.

Some may be arguing about what happens to the air several wing chord lengths away from the wing and how ground effect does have an influence on lift, while others are simply concerned with what happens near the surface of the wing which affects the wing and is involved in the creation of lift.

So while you and others are correct to a large degree that there little translation of the air in the larger vicinity of the aircraft, the more local effects near the surface of the wing need to be considered differently.

I believe that near the surface, let's say within a box 1/4 chord length on edge, the pressure regions create near vertical forces resulting in lift and the vertical acceleration and translation of the air. What happens outside of the box has less impact on the creation of lift and on the aircraft, especially since an aircraft is moving away from the air affected by the process. (This of course may not be true of a helicopter).

greg
ciurpita is offline Find More Posts by ciurpita
Reply With Quote
Old Nov 16, 2012, 07:35 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
Quote:
Originally Posted by ciurpita View Post
... on what scale?
Pick any scale you want. If the string of boxcars forms a closed loop, you can't change it's linear momentum.

Perhaps a better example... No matter how hard to push/spin a merry-go-round, you'll never be able to change it's linear momentum (assuming is has an axisymmetric mass distribution). The point I'm trying to make is you can push on something, and even put it into motion without changing its (linear) momentum.

Why is this important? Many folks here have suggested that if a lifting wing pushes on the air, then the air's momentum must be changing at a rate equal to the lift. This is not necessarily true (it's certainly not required by Newton''s Laws).
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 16, 2012, 07:36 AM
An itch?. Scratch build.
eflightray's Avatar
South Wales U.K.
Joined Mar 2003
13,179 Posts
Can the threads title be reversed ?, 'You can't have downwash without lift'.

Or, 'You can't have lift without turbulence'.

Sorry, I know, I shouldn't really be here
eflightray is offline Find More Posts by eflightray
Reply With Quote
Old Nov 16, 2012, 07:39 AM
greg
ciurpita's Avatar
somerset, nj
Joined Feb 2005
371 Posts
Quote:
Originally Posted by ShoeDLG View Post
Perhaps a better example... No matter how hard to push/spin a merry-go-round, you'll never be able to change it's linear momentum (assuming is has an axisymmetric mass distribution). The point I'm trying to make is you can push on something, and even put it into motion without changing its (linear) momentum.
yes, the merry-go-round is not changing it's linear momentum

but what about a person on the merry go round? with each spin, that person is in the same location. But what does that person feel?

... on any scale?
ciurpita is offline Find More Posts by ciurpita
Reply With Quote
Old Nov 16, 2012, 08:33 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
Not sure what you're getting at. The merry-go-round example illustrates the following:

-You exert a force on the the merry-go-round
-The merry-go-round pushes back with equal/opposite force
-The merry-go-round spins faster as a result of the force you exert
-The linear momentum of the merry-go-round isn't changing

Conlcusion: it is possible to push against an object, and put that object into motion without changing its linear momentum. I'm not sure why it matters what a person on the merry-go-round feels (besides sick).
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 16, 2012, 09:05 AM
greg
ciurpita's Avatar
somerset, nj
Joined Feb 2005
371 Posts
Quote:
Originally Posted by ShoeDLG View Post
Conlcusion: it is possible to push against an object, and put that object into motion without changing its linear momentum. I'm not sure why it matters what a person on the merry-go-round feels (besides sick).
yes, you've made your point, and i believe it is correct. A force can be applied to a mass, causing it to rotate without changing its linear momentum.

why is this important when discussing the mechanism that causes lift?
ciurpita is offline Find More Posts by ciurpita
Reply With Quote
Old Nov 16, 2012, 09:54 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
802 Posts
Quote:
Originally Posted by ciurpita View Post
why is this important when discussing the mechanism that causes lift?
Many posts in this thread have invoked Newton's Laws to support the idea that by pushing on the air, a lifting wing must change the net (linear) momentum of the air. The implication is that a wing creates lift in fundamentally the same way that a rocket creates thrust (by imparting momentum to its propellant). If that were true, you'd be able to count up the vertical mometum in the airplane's wake and show that it's rate of change is equal to the lift. That isn't the case.

Why is this important? I suppose it only makes a difference if you care to undertand that there are subtle differences between how rockets (and even jets or props) create thrust and how a wing creates lift.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 16, 2012, 10:02 AM
greg
ciurpita's Avatar
somerset, nj
Joined Feb 2005
371 Posts
Quote:
Originally Posted by ShoeDLG View Post
Many posts in this thread have invoked Newton's Laws to support the idea that by pushing on the air, a lifting wing must change the net (linear) momentum of the air. The implication is that a wing creates lift in fundamentally the same way that a rocket creates thrust (by imparting momentum to its propellant). If that were true, you'd be able to count up the vertical mometum in the airplane's wake and show that it's rate of change is equal to the lift. That isn't the case.
ok, linear momentum is not changing. The air isn't being pulled from above and forced into the ground.

is the angular momentum of the air changing? (maybe i'm wrong in calling this angular momentum, i don't want to get caught up in semantics, hopefully you understand what i mean by this)

If so, what causes it to change?
ciurpita is offline Find More Posts by ciurpita
Reply With Quote
Reply


Thread Tools

Similar Threads
Category Thread Thread Starter Forum Replies Last Post
Discussion Dymond D47 servo, Which version do you have on hand, how can you tell??? TDboy Hand Launch 5 Apr 21, 2012 02:26 AM
Discussion Is there anything listed that you can't live without? P-51C Life, The Universe, and Politics 44 Aug 05, 2011 02:55 PM
Discussion Fast food that you just can't do without. Bilbobaker Life, The Universe, and Politics 101 Aug 17, 2010 11:30 AM
Joke What Would You do to get that Aircraft You Can't Have ? Chophop Humor 10 Feb 09, 2010 07:00 AM
Discussion HELPIf you have a problem that can't be solved here ccowboyearl Beginner Training Area (Heli-Electric) 2 May 14, 2009 12:31 PM