Is the equation Velocity = Distance * Trigonometric Function valid in physics?

[Raf]

Hello,
It has been a long time since I first looked at this, so thought I might ask for some help in clarifying this problem:

Is an equation of the form --> Velocity = (Distance) * (Trigonometric function) a valid one in physics?

If so, what is the relationship of trigonometric functions and dimensions in physics in this case? It seems to me that the above relationship or equation would appear to be invalid if the trigonometric function value itself is dimensionless.

--> (L/t) ≠ L * (Dimensionless number) --> where: L is a Dimension of length and t is a dimension of time

The origin of the question comes from a proof of Hartmann's construction used to find the center of curvature of a path in kinematics.
Thank you in advance for you help.
Rafael

 

[PeroK]

Normally when you have s trig function it is either a) a function of an angle, hence dimension-less; or b) a function of the product of frequency and time,which is dimension-less.

E.g. in simple harmonic motion you may have:

$x = x_0 \cos (\omega t)$

 

[Doc Al]

Is an equation of the form --> Velocity = (Distance) * (Trigonometric function) a valid one in physics?

I would say no. (Agreeing with @PeroK )

If you could give more context as to where you've seen this used, perhaps we can see what was meant.

 

[RPinPA]

Summary: Is this equation or expression a valid one in physics?

Hello,
It has been a long time since I first looked at this, so thought I might ask for some help in clarifying this problem:

Is an equation of the form --> Velocity = (Distance) * (Trigonometric function) a valid one in physics?

No it is not. Trig functions and their arguments are dimensionless.

If so, what is the relationship of trigonometric functions and dimensions in physics in this case? It seems to me that the above relationship or equation would appear to be invalid if the trigonometric function value itself is dimensionless.

Correct.

--> (L/t) ≠ L * (Dimensionless number) --> where: L is a Dimension of length and t is a dimension of time

The origin of the question comes from a proof of Hartmann's construction used to find the center of curvature of a path in kinematics.
Thank you in advance for you help.
Rafael

Hmm. Something special is being done with units then. Perhaps there is a constant of proportionality with the correct units to make that a sensible statement, but with the numerical value of 1.

 

[Raf]

I would say no. (Agreeing with @PeroK )

If you could give more context as to where you've seen this used, perhaps we can see what was meant.

The equation in question is used in a proof (or explanation of a procedure). If you have access to a kinematics and dynamics of machines text, you will find it under "Hartmann's Construction" (look for "gauge line" and "gauge angle" part). It is not particularly long or complicated, but it seemed to me it would be too long to properly specify in a forum, so the question I posed was the best I could come with in trying to get to the kernel of it.
Hartmann's construction is used to find the center of curvature of a path of any point of a moving body graphically. It uses mostly geometry and trigonometry, but if forms part of kinematic analysis, so velocities and displacements are involved. I can mechanically perform the operation and find the center of curvature, but it has been a long time since I was introduced to the method and looking back at my old notes, I found nothing relating to the problem, it is only now that the perceived problem came to me, it seems. I don't doubt the validity of the proof of course, but at a simple glance, something seems odd about it. The best interpretation I can come up with is that if the trigonometric function relates sides of a triangle whose sides belong to different dimensions (like in the procedure an adjacent side is a magnitude of a tangential velocity vector and opposite side is length of a radius of rotation) then the ratio of both magnitudes as used in a tangent function would have the correct "units" or "dimensions" when used in the equation: velocity = distance*trigonometric function. It just seems odd when looked at in terms of trig functions and angles, where I've always understood them to be treated as dimensionless in equations of physics.
Sorry if it seems so convoluted and muddy, and many thanks again for your help.

 

[Ibix]

There's discussion of Hartmann's Construction at Google Books. It does indeed appear to draw velocity vectors on the same plane as displacement vectors, and happily construct triangles from the resulting lines and you do end up with an equation that looks like $v=d\sin\theta$.

I think it works by a horrible abuse of notation (even by physicists' rather slapdash standards). Essentially a velocity of 1m/s is assumed to be drawn as an arrow 1m long, without comment. Formally, they should specify that the velocity vectors are to be drawn at a scale if 1cm:1m/s (or whatever). Then you would be explicitly working with distance everywhere on the diagram.

However, I think the maths works out the same whatever scale factor is used, so they've omitted that formal step and told you to draw a velocity vector, implicitly expecting you to use a scale factor of 1 distance unit:1 speed unit, and hoped you wouldn't ask questions.

Well done for asking questions.

Yes, they're doing something wrong. No, it doesn't matter in this particular case (if I've followed correctly).

 

[Raf]

Ibix,
First, thanks for your reply.

It does indeed appear to draw velocity vectors on the same plane as displacement vectors, and happily construct triangles from the resulting lines and you do end up with an equation that looks like $v=d\sin\theta$.

You got to the kernel of it right away. I did not articulate it that way originally as I felt it would confuse whoever was reading it "cold turkey", so thanks for cleaning the question up.

I think it works by a horrible abuse of notation (even by physicists' rather slapdash standards). Essentially a velocity of 1m/s is assumed to be drawn as an arrow 1m long, without comment. Formally, they should specify that the velocity vectors are to be drawn at a scale if 1cm:1m/s (or whatever).

The book I used was my old Kinematics and Dynamics of Machines textbook (George H. Martin, Second Edition, McGraw-Hill). By the chapter where you come across the procedure/proof, it is already established that a suitable scale is assumed for the lengths of all dimensions to be used when plotting the geometries and vectorial quantities developed in the procedures/proofs (it's implicit, without having to mention this every time throughout the book); so this is not a fault of the authors, but rather my failure, when I fail to relate this fact with an explanation of the question I make.

Then you would be explicitly working with distance everywhere on the diagram.

As you put it, this sounds like the core of the problem, but still I am not able to "digest it"; what insight am I missing? I can see that the tangential velocity vector will always be perpendicular to the radius of rotation at that instant in time (basic physics). I can actually make a plot of this in paper (under suitable scales for the velocity units and length of the radius units) and construct a right triangle with a T and a 90 ruler that reflects this (or in CAD for that matter). What I don't see is how this is meaningful from a physics point of view as it seems the triangle itself is built by mixing apples and oranges, sort of speak ("...draw velocity vectors on the same plane as displacement vectors, and happily construct triangles from the resulting lines you do end up with an equation that looks like v=dsinθ")

However, I think the maths works out the same whatever scale factor is used, so they've omitted that formal ste and told you to draw a velocity vector, implicitly expecting you to use a scale factor of 1 distance unit:1 speed unit, and hoped you wouldn't ask questions.

I agree that the math (or geometry) part would work regardless of scale factor, but the physics part of it (if this makes sense to you), where you end with a seemingly invalid expression, is what is bugging me.

Many thanks again for your help. I hope the light bulb goes on in my head at some point

 

[Baluncore]

It just seems odd when looked at in terms of trig functions and angles, where I've always understood them to be treated as dimensionless in equations of physics.

The trigonometric functions can be represented as polynomial approximations. Any polynomial function of a dimensioned number requires each coefficient be assigned different dimensions, so that the evaluated terms will have identical dimensions before they are summed. We usually ignore the dimensions of the coefficients because it is easier to assume everything in a polynomial world is dimensionless. It is the things we ignore or throw away that come back to bite us.

 

[Ibix]

Ok - I think I understand what they're doing.

Suppose I had a clock with a pair of fireflies perched on the minute hand, and I took two photos of it in a dark room five minutes apart. I expect you're comfortable with the idea that you could locate the centre of rotation by drawing lines through the fireflies and finding the intersection.

But for a more complex machine, the centre of rotation is moving and we can't wait five minutes between photos because the lines we draw through the fireflies will pass through different centres. However, if we took two photos a tiny fraction of a second apart then (if we're precise enough) we could use the method anyway, since the centre of rotation wouldn't have moved much. And we're going to want a computer to do this, so we'll need to convert the line-drawing method into algebra. The good news is that because the rotation has been so small we can use a small angle approximation and just draw the distances the fireflies move as straight lines of length $v_i\delta t$, where $v_i$ is firefly $i$'s velocity and $\delta t$ is the time between photos. So instead of drawing the fireflies' paths as very narrow pie slices we draw very narrow triangles.

Grinding through the maths, though, we find that the $\delta t$ drops out. So the algebra is the same if we draw the fireflies' paths as lines that are $Av_i$ long, where $A$ is some constant (might as well be 1) with units of time. And the triangles are easier to draw.

So - if I'm understanding correctly - everything they're doing is valid physics as far as simplifying the maths by making a small angle approximation. Scaling up some lengths but not others, though, is not valid physics - doubly so if you quietly pull units into your scale factor and then forget about them. The maths works out in this case, though, and the process for drawing the diagram is easier than drawing and labelling infinitesimally wide triangles.

Does that make sense? The process as written is mathematically legit but not physically so, but the maths maps onto a valid bit of physics.

 

[Dr.D]

In kinematics of machines, just as everywhere else, trig functions should be considered to be dimensionless and the entire equation should be dimensionally consistent. This typically requires that the coefficient of the trig function have the necessary units to make the dimensions work correctly.