Formula for lift: Question
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Deemar, very well thought out posts. The way i think of it is that the airflow follows the foil section by the coanda effect, which induces a rotation about the foil. The rotation component then causes the bernoulli lift, while the resulting downwash over the foil causes the induced drag. In simpler language, to paraphrase Nick: the airfoil experiences lift because it is throwing air at the ground, and drag because it is slowing it up.
NASA site on Bernoulli vs Newton
i4iq, this is said tongue in cheek - i sincerely wish you a Merry Xmas:
Why can't you allow those of us who believe that the Earth is 4.5 bn years old to practice our beliefs? Having studied relativity, and being about to embark on quantum physics, then to paraphrase Shakespeare: there are more things in heaven and earth, i4iq, than are dreamt of in your philosophy.
Mart
NASA site on Bernoulli vs Newton
i4iq, this is said tongue in cheek - i sincerely wish you a Merry Xmas:
Why can't you allow those of us who believe that the Earth is 4.5 bn years old to practice our beliefs? Having studied relativity, and being about to embark on quantum physics, then to paraphrase Shakespeare: there are more things in heaven and earth, i4iq, than are dreamt of in your philosophy.
Mart
Last edited by Graviman; 29th Dec 2006 at 06:17. Reason: Inserting link.
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Hey Mart
(just as Deemar was getting this back on track...)
Thanks Mart, you too. I enjoy reading your posts!
Yes, you must feel under an enormous amount of pressure to believe!
I look forward to discussions and theories on gravitational time dilation, event horizons, dark matter (or not!) and other interesting stuff - but maybe not on this forum!
Mart, if you mean more than I can dream, yes of course - more than the philosophy itself, then logically, thats... illogical!
More than happy to hear your thoughts via PM.
Happy New Year!
i4iq
(just as Deemar was getting this back on track...)
More than happy to hear your thoughts via PM.
Happy New Year!
i4iq
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Can Some one here help me with this one..
I dont mean to fan any fire. but ..... Hang on. Fan..... what is the by product of lift ??? .... Hmmmmm.
whats that Law.. action reaction..
So would the 1/2 roh be because to put it to lift. we can only ever yeild half of the Potential.
IE. half for lift. and half for reaction. IE You go up. air goes down..
this is bugging me.
I do think either I dont understand Bernoulli theorem or it is wrong (Not completly) and looking at condra is helping.. (I think friction is the cause for the Pressure difference (IE any pressure sencor in the streem will have the Airmolicules removed from their pocket via friction and Attraction to the moveing streem..
Ne way. Ignore that last paragraph. it is just a little
during my Slow journy to become 
towards Aeordynamics..
Oh. and is there a relation to Electricty. IE Electric fields can be influenced/generated by Magnetic/electric fields.. (these things all seem similar to me, again at this early stage of my journy to becoming a)
I dont mean to fan any fire. but ..... Hang on. Fan..... what is the by product of lift ??? .... Hmmmmm.
whats that Law.. action reaction..
So would the 1/2 roh be because to put it to lift. we can only ever yeild half of the Potential.
IE. half for lift. and half for reaction. IE You go up. air goes down..
this is bugging me.
I do think either I dont understand Bernoulli theorem or it is wrong (Not completly) and looking at condra is helping.. (I think friction is the cause for the Pressure difference (IE any pressure sencor in the streem will have the Airmolicules removed from their pocket via friction and Attraction to the moveing streem..
Ne way. Ignore that last paragraph. it is just a little
during my Slow journy to become towards Aeordynamics..Oh. and is there a relation to Electricty. IE Electric fields can be influenced/generated by Magnetic/electric fields.. (these things all seem similar to me, again at this early stage of my journy to becoming a
)
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BGRing,
You have it wrong, the 1/2 x Rho x Vsquared is derived by integrating the rate of change of the momentum. Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x Vsquared.
It is not that half of anything is lost, it is that the area under the momentum vs V curve is calculated by the formula.
You have it wrong, the 1/2 x Rho x Vsquared is derived by integrating the rate of change of the momentum. Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x Vsquared.
It is not that half of anything is lost, it is that the area under the momentum vs V curve is calculated by the formula.
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1/2? x Rhox (V squared) is derived by integrating the rate of change of the momentum.
Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x (V squared).
it is that the area? under the momentum vs V curve is calculated by the formula.
The Purple and Greay parts confuse me. (Gentle reminder, Why 1/2)
Please remember I want to be but curently I am 
so could you dumb it down a little(Lot) for me.
I may well be a Lost cause 
Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x (V squared).
it is that the area? under the momentum vs V curve is calculated by the formula.
The Purple and Greay parts confuse me. (Gentle reminder, Why 1/2)
Please remember I want to be
but curently I am so could you dumb it down a little(Lot) for me.
Edit 1/2 hour later
Or is it
1/2 x Rho x Vsquared is derived by integrating the rate of change of the momentum.
Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x Vsquared.
it is that the area? under the momentum vs V curve? is calculated by the formula.
Changes being the Green bits
end Edit
Or is it
1/2 x Rho x Vsquared is derived by integrating the rate of change of the momentum.
Since the momentum is Rho x V, if you integreate it (using integral calculus) you get the 1/2 x Rho x Vsquared.
it is that the area? under the momentum vs V curve? is calculated by the formula.
Changes being the Green bits
end Edit
Last edited by BGRing; 17th Jul 2007 at 11:11.
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I don't whether this will make things better or worse but ....
Differential calculus is what Nick is talking about.
Differentiation is how one property changes with another and integration (i.e. integral calculus) is the reverse. I'll use a familiar example.
If we take the rate of change of distance with time (i.e. we differentiate distance over time) we get speed (i.e. Speed = distance/time). If we differentiate speed with time, we get acceleration (i.e. acceleration = speed/time and we know speed = distance over time) which equals distance/(time squared).
Therefore, if we integrate acceleration, we get speed and if we integrate speed, we get distance.
If you draw a graph with time along the x axis (horizontal) and speed along the y-axis, the area under the graph will be the distance covered. If the speed is constant, the the graph will be a horizontal line. However, the speed could vary but the area under the graph will be the distance. If the speed is uniformally increasing, then there will be a slope to the graph and this slope is the acceleration.
As a rough schoolgirl guide, I can remember that you integrate, you add one to the power and divide by the power (and add a constant!!). To differentiate, multiply by the power and take one off the power.
i.e. momentum = mass x velocity. This means velocity has a power of one. Therefore, to integrate it by adding one to the power means velocity is to the power of two i.e. squared. Divide by the power i.e. divide by 2.
This sort of maths (which I have dragged up from over 25 years ago) is best shown rather than read so if you can find a friendly maths teacher, it would help you no end. It is easy honest but not to learn by yourself.
Cheers
Whirls
Differential calculus is what Nick is talking about.
Differentiation is how one property changes with another and integration (i.e. integral calculus) is the reverse. I'll use a familiar example.
If we take the rate of change of distance with time (i.e. we differentiate distance over time) we get speed (i.e. Speed = distance/time). If we differentiate speed with time, we get acceleration (i.e. acceleration = speed/time and we know speed = distance over time) which equals distance/(time squared).
Therefore, if we integrate acceleration, we get speed and if we integrate speed, we get distance.
If you draw a graph with time along the x axis (horizontal) and speed along the y-axis, the area under the graph will be the distance covered. If the speed is constant, the the graph will be a horizontal line. However, the speed could vary but the area under the graph will be the distance. If the speed is uniformally increasing, then there will be a slope to the graph and this slope is the acceleration.
As a rough schoolgirl guide, I can remember that you integrate, you add one to the power and divide by the power (and add a constant!!). To differentiate, multiply by the power and take one off the power.
i.e. momentum = mass x velocity. This means velocity has a power of one. Therefore, to integrate it by adding one to the power means velocity is to the power of two i.e. squared. Divide by the power i.e. divide by 2.
This sort of maths (which I have dragged up from over 25 years ago) is best shown rather than read so if you can find a friendly maths teacher, it would help you no end. It is easy honest but not to learn by yourself.
Cheers
Whirls
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seeing as you are online. I will type a quick reply.
Thanks for the Reply and effort.
I ahavent read it yet. and I dont Imagin I will understand it first time round.
Will get back with my findings. (if that made things click for me)
Thanks again.
OK. reading it now (May take 10 min
. may take 10 days )
Thanks for the Reply and effort.
I ahavent read it yet. and I dont Imagin I will understand it first time round.
Will get back with my findings. (if that made things click for me)
Thanks again.
OK. reading it now (May take 10 min
. may take 10 days )
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National Geographics 2006 Darwin & Evolution
"When scientists say a "theory," they mean a statement based on observation or experimentation that explains facets of the observable so well that it becomes accepted as fact. They do not mean an idea created out of thin air, nor do they mean an unsubstantiated belief."
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What's that got to do with the price of eggs?
How would you reconcile Einstein's Theory of General Relativity and the Theory of Quantum Mechanics as postulated by Heisenberg and Schrodinger? Neither or both are fact as they essentially contradict each other!!
Schrodinger's Cat may (or may not) disappear into thin air (depending on whether you look in the box) and Heisenberg's Uncertainty Principle means you can never be sure where the cat is. Pauli's Exclusion Principle means that no two cats are ever in the same position or state of excitement.
Let's face it, some of these ideas pretty much came out of thin air (or maybe mescalin induced) and are very, very hard to substantiate.
Cheers
Whirls
How would you reconcile Einstein's Theory of General Relativity and the Theory of Quantum Mechanics as postulated by Heisenberg and Schrodinger? Neither or both are fact as they essentially contradict each other!!
Schrodinger's Cat may (or may not) disappear into thin air (depending on whether you look in the box) and Heisenberg's Uncertainty Principle means you can never be sure where the cat is. Pauli's Exclusion Principle means that no two cats are ever in the same position or state of excitement.
Let's face it, some of these ideas pretty much came out of thin air (or maybe mescalin induced) and are very, very hard to substantiate.
Cheers
Whirls
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Davey. I think I know what You mean
My keyboard has no numpad so I can not do Alt codes.
as I had put it above 1/2 x roh x VSquared would work on it own. But that is not the formula so if the whole formula was to be writen this way it would come out incorect. thus the (brackets)
Lift = Cl x (roh x 0.5) x (VxV) x S
would that work or would you still need to add the [] for the roh portion. As I think i understand it. the brackets are to instruce you to use the answer to the math inside them.
where if it were not in brackets then CL would be halved and then Times roh (being the incorect method)
WirlyGig. You are way smarter than I could ever be (Due mainly to my Interests and needs) and I have decided that it is not all bad to be a bit of a
. My brain cringes when I see the formula you had in the last link. I understood you Example about Distance over time is Speed and Speed variations over time give Aceleration differences and that the Area under them are Speed and Aceleration respectivly. I think I understood how you used time twice thus Squared it. but still dont know where you halved a Constant..
not to worry. I have a friend who i a math teacher. He understood the examples given, though due to is relativly lesser experience with Aerodynamics and Physics. was not (Just yet) able to describe the lift formula. though , he did say that the 1/2 goes with the (VxV).. So I will enlist his abilitys to bring my Relativly Slow Brain, and its comprehension, up to date.
Thanks to you all for your help.
PS. Nick. My heli book says it is an average of the start and End velocity. I couldn't see it due to the placement of it at the r instead of after (VxV) x 0.5 x S . and because the book told me to refer to Dynamic energy.
and in this book. it says
Dynamic Energy is (1/2xM) x (VxV)
thinking on it.
the total Energy spent is Momentum. IE MxV
but that is the end result. not the average of both.
My brain hurts.. I still see that it is because you are minusing the action from momentum to get reaction. THats it. I am lost..
Gona call the Math teacher now.
My keyboard has no numpad so I can not do Alt codes.
as I had put it above 1/2 x roh x VSquared would work on it own. But that is not the formula so if the whole formula was to be writen this way it would come out incorect. thus the (brackets)
Lift = Cl x (roh x 0.5) x (VxV) x S
would that work or would you still need to add the [] for the roh portion. As I think i understand it. the brackets are to instruce you to use the answer to the math inside them.
where if it were not in brackets then CL would be halved and then Times roh (being the incorect method)
WirlyGig. You are way smarter than I could ever be (Due mainly to my Interests and needs) and I have decided that it is not all bad to be a bit of a
. My brain cringes when I see the formula you had in the last link. I understood you Example about Distance over time is Speed and Speed variations over time give Aceleration differences and that the Area under them are Speed and Aceleration respectivly. I think I understood how you used time twice thus Squared it. but still dont know where you halved a Constant.. not to worry. I have a friend who i a math teacher. He understood the examples given, though due to is relativly lesser experience with Aerodynamics and Physics. was not (Just yet) able to describe the lift formula. though , he did say that the 1/2 goes with the (VxV).. So I will enlist his abilitys to bring my Relativly Slow Brain, and its comprehension, up to date.
Thanks to you all for your help.
PS. Nick. My heli book says it is an average of the start and End velocity. I couldn't see it due to the placement of it at the r instead of after (VxV) x 0.5 x S . and because the book told me to refer to Dynamic energy.
and in this book. it says
Dynamic Energy is (1/2xM) x (VxV)
thinking on it.
the total Energy spent is Momentum. IE MxV
but that is the end result. not the average of both.
My brain hurts.. I still see that it is because you are minusing the action from momentum to get reaction. THats it. I am lost..
Gona call the Math teacher now.
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Thanks for the link. That thing has me buged also.
I dont dispute the Phenomenon. just the Explanation of what is happening.
untill recently I would have said something like the following;
But recently. I am Leaning toward just accepting the Pressure difference Explanation. well almost
Seen wiki and nasa and plenty other sites. seen Bernoulli water device and it all seams to be explainable in another way to me. Look at the Scram jet. dose air pressure drop at the Point of Squish ? ... More investigating by be and some more reprograming of my current Physics modle inside my head may be needed
Oh. I was good at math. but only when I understand the Reason for performing the calculations. (Never did calculus or Dont remeber it, Never did Physics either... and kinda did bad in English. and My spelling never improved either. But ... Spelling and Math is not everything. though they can be applyed to everything : )
I dont dispute the Phenomenon. just the Explanation of what is happening.
untill recently I would have said something like the following;
has anyone ever tested the Airpressure measured as a particle traveling through said Venturi. (IE not from a Port on the wall of the tube. where the port could have particle removed from it by the stream)
IE
have a Pressure guage on a neutraly boyant ballon float through the tube (Increase the length of tube) same for Water.
Personaly. I am Not beleiving the pressure difference.
I belive there is Equality in the tube. (Its is relativly neutral inside, as a ballon floats through it is relative to the air, without dynamic energy)
Show me a Specific example of why you think there is a pressure difference. and I could explain (Without math) why it appears that way.
friction and attraction with a little bumping in general direction. think of these when you look at every example of the measuring methods of a Venturi. (Thing is it works. IE the Venturi tube dose draw air liquid etc into the stream. but I'm not sure it is pressure difference.
IE
have a Pressure guage on a neutraly boyant ballon float through the tube (Increase the length of tube) same for Water.
Personaly. I am Not beleiving the pressure difference.
I belive there is Equality in the tube. (Its is relativly neutral inside, as a ballon floats through it is relative to the air, without dynamic energy)
Show me a Specific example of why you think there is a pressure difference. and I could explain (Without math) why it appears that way.
friction and attraction with a little bumping in general direction. think of these when you look at every example of the measuring methods of a Venturi. (Thing is it works. IE the Venturi tube dose draw air liquid etc into the stream. but I'm not sure it is pressure difference.
Seen wiki and nasa and plenty other sites. seen Bernoulli water device and it all seams to be explainable in another way to me. Look at the Scram jet. dose air pressure drop at the Point of Squish ? ... More investigating by be and some more reprograming of my current Physics modle inside my head may be needed
Oh. I was good at math. but only when I understand the Reason for performing the calculations. (Never did calculus or Dont remeber it, Never did Physics either... and kinda did bad in English. and My spelling never improved either. But ... Spelling and Math is not everything. though they can be applyed to everything : )
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Don't worry about the 1/2. Coefficients in an equation are there just to give you a number, not really to give you understanding. The 1/2 could just as easily be embedded into the Cl, although then some other equations would need a factor of 2 included.
The funny thing is that the 1/2 is typically kept in the equation to aid the understanding. As has been mentioned before, that number just falls out of the mathematics and not the physics. And here we are concerned with the physics, but wondering about the 1/2. Be much more concerned with the exponent on the airspeed term...it actually matters.
As far as having a different understanding than Bernoulli, don't be too concerned. These theories do not define how nature behaves, they are just an attempt at quantifying how nature behaves. You can describe many systems using different physical concepts without revealing an inconsistency. An example of this is in basic kinematics, describing the motion of an unaccelerated particle. You can give its mass and momentum, its velocity and momentum, its velocity and energy, its momentum and energy, etc. Just because you're using different terms and concepts, doesn't mean there is anything different about nature.
Bernoulli's theorem is consistent with everything known about classical mechanics (for fluid dynamics, perhaps not for airfoils). You can define static and dynamic pressure, and then demonstrate that static pressure in a moving fluid decreases with the velocity of the flow. You can then demonstrate it experimentally. If you want to describe it using concepts other than static and dynamic pressure, go ahead. Doesn't mean that Bernoulli was wrong.
The funny thing is that the 1/2 is typically kept in the equation to aid the understanding. As has been mentioned before, that number just falls out of the mathematics and not the physics. And here we are concerned with the physics, but wondering about the 1/2. Be much more concerned with the exponent on the airspeed term...it actually matters.
As far as having a different understanding than Bernoulli, don't be too concerned. These theories do not define how nature behaves, they are just an attempt at quantifying how nature behaves. You can describe many systems using different physical concepts without revealing an inconsistency. An example of this is in basic kinematics, describing the motion of an unaccelerated particle. You can give its mass and momentum, its velocity and momentum, its velocity and energy, its momentum and energy, etc. Just because you're using different terms and concepts, doesn't mean there is anything different about nature.
Bernoulli's theorem is consistent with everything known about classical mechanics (for fluid dynamics, perhaps not for airfoils). You can define static and dynamic pressure, and then demonstrate that static pressure in a moving fluid decreases with the velocity of the flow. You can then demonstrate it experimentally. If you want to describe it using concepts other than static and dynamic pressure, go ahead. Doesn't mean that Bernoulli was wrong.
