Endercreeper01

 

I don't understand your equivocation. The entire article at the link goes further into the mathematical constraints/techniques at play, and if that doesn't help you then it doesn't help yo

It doesn't help very much. I wanted a list of factors.

I'm no expert but I have recently had occasion to do some studying on the lift coefficient at some pages @ NASA. I found them very informative and perhaps you'll find them helpful as well.

 

The Lift Coefficient

I don't mean based on the lift force. I mean the factors that affect the coefficient of lift itself

What are the factors affecting the coefficient of l.ift?

The only thing I see as debatable is should we change the standard. i.e. should we teach kids that a circle has τ radians instead of 2π radians.

 

Other than that it is just a non-issue. It's not as if we are trying to redefine 2π→π or anything...

π and τ are two different symbols, so there's no reason we can't even use them both interchangeably.. You don't have to pick one.

It's like getting into a heated argument over whether we should use h or h-bar. Just use which ever one you prefer in any specific situation, or even together if you're feeling crazy enough (like using τπ instead of 2π²)

 

As far as the question "should we change the school standard?" goes, I'd ask why this one tiny little change should be more important than any other standard that seems off due to historical reasons. For example, we still talk about "current" as positive charge flow, although talking about anything moving in a current would be much more accurate if it had been defined as negative (or even better, if electrons had just been defined as positive). But it works, so we don't bother changing it.

As long as everyone knows what's going on and they are free to define new things, or redefine things as they see fit in any work or papers they might write.

If you want to define current as negative charge flow nobody will stop you as long as you put 1 line in your paper saying you're doing so, same with τ.

We are arguing if pi or tau is the better circle constant

Remember also that for small angles sinx = tanx = x.

how would sinx=tanx=x with small angles? tanx=sinx/cosx

And also check out this link: www.tauday.org

Yes they do. If a photon collides with the blocking object, it collides according to everyone.

 

Collisions are absolute.

But the size of the object changes relative to different observers

I think your understanding of fluid dynamics is insufficient for you to postulate theories on fluid dynamics...

How is it insufficient?

So what do you think?

Only if you're making up some new definition for Re.....Feel free to imagine whatever you want.

Re=Dvρ/μ according to my theory then they must have different reynolds numbers and also what would θ be for a half-sphere?

I think it is

45. Is that correct? According to my theory, a half sphere and sphere have the same Cd. And also i found this link: https://www.google.com/url?sa=t&source=web&cd=2&ved=0CBIQFjAB&url=http%3A%2F%2Fwww.chem.mtu.edu%2F~fmorriso%2FDataCorrelationForSphereDrag2013.pdf&ei=V0QtUuChBNb94APqo4HICQ&usg=AFQjCNFux5E9ERsURhBQxTL8Zq5U-KqChw&sig2=Abqy5fRUhntiDepifZMMjQ it talks about Cd based on Re up to Re=10⁷ so

If i divide that by the cosine of 45, which is 2^1/2/2 then i get how to calculate the coefficient of drag for a plate up to Re=10⁷. So the equation becomes 48/2^1/2 Re + 5.2Re/5)/2^1/2(1+(Re/5)^1.52 + .822(Re/263000)^-7.94/2^1/2(1+(Re/263000)⁸)+ 2Re^0.8/463000*2^1/2

Ummmmmmmmmmmmm......NO!

 

Reynolds numbers are only dependent on the flow, not the shape of any submerged bodies. Turbulence, which is represented by reynolds number, has an effect on drag and that's why the drag coefficient is not a constant.

I know that, but according to my theory, they do. How can you be sure they don't? Its likely that they all have different Re

Yes they do. If a photon collides with the blocking object, it collides according to everyone.

 

Collisions are absolute.

Yes, but then the object shrinks relative to an outside observer by a factor of gamma and so less light relative to them

Good start, now try the other shapes.

The shapes all have roughly the same Reynolds numbers. The sphere and half sphere both have the same average angle and roughly the same Reynolds number, and so coefficients that are very similar. The same goes for the streamlined half- and full bodies. I don't know about the long and short cylinder because they must have very different Re, and I don't know the angle for the cone.

Well then, perhaps you should seek out an imaginary forum to seek out imaginary answers to your imaginary problem. Then again, since it's imaginary you could just make up the answers that you want.

I cant just make the answers what I want, I have to use theory. It is 2 dimensional, and the closer that angle is to 90 degrees, then the less drag coefficient there is. It then must be dcosθ and d is the drag coefficient it would have at 0 degrees.

Are you saying you are talking about an imaginary plate? All plates in the real world are 3 dimensional.

That imaginary plate is the basis for my theory

 

 

 

First, which one is it? Relative to the x axis or not? First you say it is then you say it's not.

 

Secondly, a submerged plate in a fluid flow still has boundary layer drag even when it is parallel to flow. See Boundary Layer at Thermopedia. Note: The coefficient of drag is referred to as the coefficient of friction in their analysis.

First, by x axis i meant axis perpendicular to direction of motion, and second, im talking about a 2 dimensional plate, not a 3 dimensional one

It's not my theory so it's not my job to test it.

That chart gives values of 1.05 and 0.8 for a cube depending on whether it's face on or edge on.

Does the proposed model also show a 30% difference?

 

If not, you can give up on it.

It it about a 30% difference in the cosine of 0 and 45 degrees, so it is correct. And according to my theory, they would have Reynolds numbers that are roughly equal

You shouldn't need one. Your stated theory from your first post says the drag coefficient is Dcosθ where θ is the angle the plate is tilted relative to the x axis. This is the same as saying that Cd varies from D ·1 to D ·0 as θ varies from 0° to 90°. This implies that the coefficient of drag would approach 0 as the angle of the plate approaches 90° relative to the x axis.

Its not relative to the x axis, its relative to the surface perpendicular to the direction of motion. Just think about it. When that angle is 90 degrees, it is in 2 dimensions, and no part of it is interacting with the air at 90 degrees, so therefore, the coefficient is zero, and the total drag is zero.

Are you suggesting the forces on a plate in a fluid stream are somewhat equivalent to those from a jet of fluid impacting on the plate?

No, because in the second case, then you would add that extra force to get the total air resistance when you want air resistance. Im talking about the coefficient of drag.

Chose an arbitrary starting value and see if the relative answers come out right.

Fine, Ill pick Re=10³. Now show me what it would be for a plate with that Reynolds number.

OK, that looks like a fair point.

Pick a Reynolds number and answer the question.

Apply your theory and see how well it matches these measured drag coefficients:

I can't find any graphs for a plate

According to the graph, it does both. And also, what do you think of my theory J.C.Maxwell?

A light source that really moves around, really sprinkles the light all around, which increases the entropy of the light. All observers agree about entropy.

Yes, but they don't agree on how much the light is blocked.

They are average measured values. As evidence by your own previous post there can be quite a bit of variability caused by turbulence. Aside from that how well does your theory match these values?

Average? it doesn't say the Reynolds number each shape had when being tested.

I agree with md - you are using bad reasoning to try and end critical thought:

You have utterly failed to demonstrate that any net benefit would accrue from such a change.

And who are you to tell me that I have to use the radius?

However, just to humour youI have used the radius in the above, where appropriate.

This is utter nonsense.

You highlighted my entire post where I offered you a selection of snap formulae containing non even multiples of pi in response to a claim that the are all even multiples and wrote something unintelligible to it.

Except for the greater part of humanity that uses the diameter because they actually have to measure this quantity.

This challenge still stands, unanswered.

mad indeed.

 

I am arguing for a balanced view.

I am also arguing for not mixing up separate issues and using one to bolster the argument for or against another.

 

I see three spearate issues here.

And you have only scratched the surface of the any of these.

 

1) Firstly, Pi is a particular number on the number line. Should we replace it with another number in formulae, equal to double its value, thus removing the need for a 2 in certain formulae.

 

2) Alternatively should we incorporate that 2 in another part of the formula?

For instance

[math]circumference = 2(\pi R)[/math]

The 2 could be applied to the pi or the R

[math] = (2\pi )®[/math]

or

[math] = (\pi )(2R) = \pi D[/math]

 

I see the diameter v radius issue as separate from the pi v 2pi issue.

Why should we specify a circle by its radius, not its diameter? We do not do this , sorry we cannot do this fro any other conic curve eg an ellipse so why is the circle special?

In fact it has been found convenient to have both diameter and radius available and to choose the most appropriate for the job in hand.

 

3) Thirdly, if we did introduce a new constant why choose tau? This symbol is already heavily overworked in many different disciplines, unlike pi which has only a few alternative uses (continuous product and Buckingham's theorem.)

I dont care what symbol we use, but for argument 2, then radius is better then diameter because a circle is defined by its radius.