It's up to you. I could have made the coefficient of dx/dt 1 and it wouldn't have really mattered too much save for a few different expressions. You can consider my choice more or less arbitrary.
We want to simplify the ODE as much as possible in the dimensionless form, so what we do is define the scaling factors (e.g. x_s, t_s) such that the ODE ends up with the fewest number of parameters possible. Now how do we 'get rid' of the parameters from the ODE (e.g. k*t_s^2/m)? Well, we define t_s such that the term k*t_s^2/m equals 1, so that the term k*t_s^2/m cancels out from the ODE, and we end up with an equation that has a smaller number of parameters (i.e. the purpose of nondimensionalization). If you need more explanation, feel free to ask!
Oh.. awesome.. I get it now. Since you have control over the scale factors, you "tweak" them in such a way that you get an equation with smaller number of parameters.. :D Thank you!
Great video! I have one question: how to go about nondimensionalizing a system of ODEs? Say, for example, a biochemical network of intracellular species where the characteristic units are time and concentration.
Thank you! Well, it's difficult to answer that question in general terms since it depends a lot on how your system and boundary/initial conditions are structured. Perhaps this video, where I nondimensionalize a complicated example system, might offer some guidance? th-cam.com/video/8CIPQNfMe5k/w-d-xo.html
But sir, how can we? I mean, how can we 'set' the co-efficients to be one? (That too twice?) K and M are properties of the system and Ts is a constant(a second in this case), then how can we relate these things?
If i understand correctly, the objective is to write the motion equation in original way. The relationship between parameters is rely on the defination of the two coefficients. Thus the motion equation are is changed.
Thank you very much. You've cited about x centered in 0, I did not understand very well and would like to know if I can always take the reference value as 0. Another question: we get a parameter with dimension inside the cossen, how to lead with this once cossen is a complex function? Many thanks once again!
What do you do if your reference terms are not 0? how do you deal with that new derivative? would it be easier if the term that wasn't zero was in the numerator or denominator?
The reference term is a constant so the dimensionless derivative will be unaffected by the reference term and everything will still work out. Hope that helps!
I don't get the part where you make each term in the ode dimensionless by dividing the entire equation by the coefficient of the second derivative. Can you explain a little more?
Are you confused about why it becomes dimensionless? Well, the variables are all dimensionless at this point so d2xtilde/dttilde2 will be dimensionless, so if I divide out the coefficient, all the individual terms become dimensionless. This is because when you're adding/equating multiple quantities, they have to have the same dimensions. Hope that helps!
Thank you for your explanation. But what if there are boundary conditions. Like x(0)=0, derivative x(0)=v0. x(0)=0 means that we can assume time starts at 0 and x is at center 0. What can I get from derivative x(0)=v0?
Well edited video. I'm still stuck on the term "dimensionless" itself. I'm missing the intuition. What dimensions did our equation have at the beginning? What dimensions did we remove? I assume all of them, hence dimensionless--without dimension. We use the term "dimension" when referring to the number of variables, usually. A line is one-dimensional, having one variable. A plane is two dimensional, and so on. I don't feel like we're using the term in the same way here. Are we?
Thank you! We aren't using the term in the same way. By dimensions, I'm referring to units. The quantity 1 meter has units of meter and dimensions of length. By converting the equation to dimensionless form, what I'm doing is making all the variables in the equation without any units. So the x^tilde is just a length ratio instead of a pure length like x. In other words, x^tilde has no units. Hope that helps!
what do you mean by potential auxiliary condition? I was taught in course that , the number of free parameters left at the end of equation would be equal to number of parameters i had started with - the number of substitution i made. But in this particular case you made two substitution ,so according to my calculation there should have been 3 free parameters would have been left at the end . But you are showing only 2. Please make it clear sir! TIA
Thank you for the suggestion! I've already showed how to nondimensionalize Navier-Stokes, which the Reynolds equation is derived from. Maybe you could have a look at that video and let me know if that helps: th-cam.com/video/8CIPQNfMe5k/w-d-xo.html
When I do the dimensionless process, after solve the non-dimensional equation then convert back to dimensional form, the results show a big error. Can you guess what is the problem?? Thank you so much
How could we choose non-dimensional variables according to the geometry. I'm always getting confusion while considering the non-dimensional variables. Could you please help me
It depends on your problem and the geometry you're dealing with. It might be more convenient to make variables dimensionless with respect to the radius but in some cases, it might be more convenient to make them dimensionless with respect to diameter. I can't really give an answer without knowing the particular problem you're solving.
The reply is quite late, hope you have found the reason :) basically its a way to simplify the expression bts/m, if you plug in your ts expression into bts/m and square the whole term, you can do some manipulation to get bsquared/mk, to get back to the original form, just take the sqrt of this, hence b/sqrt(mk).
This video did a better job at teaching nondimensionalisation than two modules at my university. Thank you.
Extremely well-explained in a quick, simple to understand manner, well done!
This is exactly what I was looking for. Thanks a lot! 🙌🙌🙌🙌
When it comes to nondimensionalization you are the one in control of your destiny.
why did you choose to make x coefficient=1 instead of dx\dt?
It's up to you. I could have made the coefficient of dx/dt 1 and it wouldn't have really mattered too much save for a few different expressions. You can consider my choice more or less arbitrary.
Thank you so much for this video!
I don't seem to be understanding why you equate those terms to 1. What does that mean ?
We want to simplify the ODE as much as possible in the dimensionless form, so what we do is define the scaling factors (e.g. x_s, t_s) such that the ODE ends up with the fewest number of parameters possible. Now how do we 'get rid' of the parameters from the ODE (e.g. k*t_s^2/m)? Well, we define t_s such that the term k*t_s^2/m equals 1, so that the term k*t_s^2/m cancels out from the ODE, and we end up with an equation that has a smaller number of parameters (i.e. the purpose of nondimensionalization).
If you need more explanation, feel free to ask!
Oh.. awesome.. I get it now.
Since you have control over the scale factors, you "tweak" them in such a way that you get an equation with smaller number of parameters.. :D
Thank you!
Exactly, and no problem!
Thank you Sir, you have made my day!
Great video! I have one question: how to go about nondimensionalizing a system of ODEs? Say, for example, a biochemical network of intracellular species where the characteristic units are time and concentration.
Thank you!
Well, it's difficult to answer that question in general terms since it depends a lot on how your system and boundary/initial conditions are structured. Perhaps this video, where I nondimensionalize a complicated example system, might offer some guidance? th-cam.com/video/8CIPQNfMe5k/w-d-xo.html
But sir, how can we? I mean, how can we 'set' the co-efficients to be one? (That too twice?) K and M are properties of the system and Ts is a constant(a second in this case), then how can we relate these things?
If i understand correctly, the objective is to write the motion equation in original way. The relationship between parameters is rely on the defination of the two coefficients. Thus the motion equation are is changed.
Thank you very much. You've cited about x centered in 0, I did not understand very well and would like to know if I can always take the reference value as 0. Another question: we get a parameter with dimension inside the cossen, how to lead with this once cossen is a complex function? Many thanks once again!
thank you so much
What do you do if your reference terms are not 0? how do you deal with that new derivative? would it be easier if the term that wasn't zero was in the numerator or denominator?
The reference term is a constant so the dimensionless derivative will be unaffected by the reference term and everything will still work out. Hope that helps!
Great video! thanks a lot.
Very helpful. Thank you so much =)
I don't get the part where you make each term in the ode dimensionless by dividing the entire equation by the coefficient of the second derivative. Can you explain a little more?
Are you confused about why it becomes dimensionless? Well, the variables are all dimensionless at this point so d2xtilde/dttilde2 will be dimensionless, so if I divide out the coefficient, all the individual terms become dimensionless. This is because when you're adding/equating multiple quantities, they have to have the same dimensions. Hope that helps!
Thanks!
Is this the same thing as putting a MSD equation into a State Space Representation, but just another name for it?
Thank you for your explanation. But what if there are boundary conditions. Like x(0)=0, derivative x(0)=v0. x(0)=0 means that we can assume time starts at 0 and x is at center 0. What can I get from derivative x(0)=v0?
I liked the explanation. Thank you.
Glad it was helpful!
Replying 7 years after the video is true commitment, respect@@FacultyofKhan
Well edited video. I'm still stuck on the term "dimensionless" itself. I'm missing the intuition. What dimensions did our equation have at the beginning? What dimensions did we remove? I assume all of them, hence dimensionless--without dimension.
We use the term "dimension" when referring to the number of variables, usually. A line is one-dimensional, having one variable. A plane is two dimensional, and so on. I don't feel like we're using the term in the same way here. Are we?
Thank you!
We aren't using the term in the same way. By dimensions, I'm referring to units. The quantity 1 meter has units of meter and dimensions of length. By converting the equation to dimensionless form, what I'm doing is making all the variables in the equation without any units. So the x^tilde is just a length ratio instead of a pure length like x. In other words, x^tilde has no units. Hope that helps!
It sure did help me. Thanks!
Thanks for the video
No problem! Glad you liked it!
what do you mean by potential auxiliary condition? I was taught in course that , the number of free parameters left at the end of equation would be equal to number of parameters i had started with - the number of substitution i made. But in this particular case you made two substitution ,so according to my calculation there should have been 3 free parameters would have been left at the end . But you are showing only 2. Please make it clear sir! TIA
Can you do the three methods of non dimensionalizing of Renolds eqn?
Thank you for the suggestion! I've already showed how to nondimensionalize Navier-Stokes, which the Reynolds equation is derived from. Maybe you could have a look at that video and let me know if that helps:
th-cam.com/video/8CIPQNfMe5k/w-d-xo.html
When I do the dimensionless process, after solve the non-dimensional equation then convert back to dimensional form, the results show a big error. Can you guess what is the problem?? Thank you so much
why you yould make kts^2/m=1?
How could we choose non-dimensional variables according to the geometry. I'm always getting confusion while considering the non-dimensional variables. Could you please help me
It depends on your problem and the geometry you're dealing with. It might be more convenient to make variables dimensionless with respect to the radius but in some cases, it might be more convenient to make them dimensionless with respect to diameter. I can't really give an answer without knowing the particular problem you're solving.
how to non diminsionalise the momentum equation
Awesome!!!!!!
CR7 of nondimensionalization! Thank you man
why b/sqrt(mk)?
The reply is quite late, hope you have found the reason :)
basically its a way to simplify the expression bts/m, if you plug in your ts expression into bts/m and square the whole term, you can do some manipulation to get bsquared/mk, to get back to the original form, just take the sqrt of this, hence b/sqrt(mk).
I'm having problem with nondimention
7:19 hard quote to end a video on lol
is there any video plz share
You can't do this. Use the chain rule it works every time.
I did use the chain rule though...