Why does Δs = rΔθ become inaccurate for large angles?

[FieldvForce]

I know how this formula is made, but surely Δs becomes highly inaccurate for large values of Δθ and r?

In fact when Δθ is ∏ surely Δθ ≠ Δs/r.

Elucidation would very appreciated.

 

[arildno]

No, if "s" is the arc length on a circle, and the angle is measured in radians, then the relations holds exactly for ALL values.

 

[PeroK]

It holds even for θ = 2∏!

 

[FieldvForce]

I know it holds for arc length but it's presenting that 's' as displacement, it's even linked linear velocity to this rule.

v = rω...how?

v = rω can't hold for 2∏ as displacement is 0...

 

[A.T.]

I know how this formula is made, but surely Δs becomes highly inaccurate for large values of Δθ and r?

That's why the correct formula uses "d" not "Δ".

v = rω can't hold for 2∏ as displacement

This formula doesn't have displacement in it.

 

[PeroK]

I know it holds for arc length but it's presenting that 's' as displacement, it's even linked linear velocity to this rule.

v = rω...how?

$Because \ v = lim_{t \rightarrow 0}\frac{ds}{dt} = lim_{t \rightarrow 0}r\frac{dθ}{dt}$

 

[Nugatory]

I know it holds for arc length but it's presenting that 's' as displacement, it's even linked linear velocity to this rule.

Even if $s$ is not the arc length, but the straight-line displacement between two points, $ds=\frac{d\theta}{r}$ is still exact because it's stated in terms of infinitesimals. If you do the line integral along the arc, you'll get $\Delta{S}=\frac{\Delta{\theta}}{r}$ where $\Delta{S}$ is the length along the arc.

 

[FieldvForce]

That's why the correct formula uses "d" not "Δ".


This formula doesn't have displacement in it.

I just used "d" because i didn't know how to use the symbol Δ in the title.

v = rω is concluded from avg(v)= r(avg)ω which is derived from using Δs/Δt = rΔθ/Δt

Yes to all posters talking about t approaching 0 and infinitesimals I agree it works there, but when t is large doesn't the equation become inaccurate?

 

[FieldvForce]

Ah, I said when ds is very big and dθ when i should have said dt.

 

[Dale]

I know how this formula is made, but surely Δs becomes highly inaccurate for large values of Δθ and r?

In fact when Δθ is ∏ surely Δθ ≠ Δs/r.

Elucidation would very appreciated.

Well, the formula in the title is incomplete.

The exact formula for distance in a plane is given by the Pathagorean theorem:
ds²=dx²+dy²

The exact transformation to polar coordinates is
x = r cos(θ)
y = r sin(θ)

Substituting the exact transformation into the exact distance you get exactly
ds² = dr² + r² dθ²

Which is the complete formula. Of course, if you want to calculate a large distance then you have to integrate.

In the special case that dr=0 then you get your formula
dθ = ds/r

 

[Chestermiller]

If Δs is a displacement vector (i.e., relative position vector) between two points on a circle, then the equation only applies infinitecimally. If Δs is the arc length between two points on the circle, then it works for all values of Δθ. Did you mean to imply that Δs is supposed to be a relative position vector?

 

[Dale]

FieldvForce, if Chestermiller is correct then you should be aware that a position vector is a path where dr≠0, so you cannot use your incomplete formula. You would need to use the full formula instead. In that sense, it would be inaccurate since you are using a simplified formula in a circumstance which violates the assumption on which the simplification is based.

 

[FieldvForce]

Well, the formula in the title is incomplete.

The exact formula for distance in a plane is given by the Pathagorean theorem:
ds²=dx²+dy²

The exact transformation to polar coordinates is
x = r cos(θ)
y = r sin(θ)

Substituting the exact transformation into the exact distance you get exactly
ds² = dr² + r² dθ²

Which is the complete formula. Of course, if you want to calculate a large distance then you have to integrate.

In the special case that dr=0 then you get your formula
dθ = ds/r

If Δs is a displacement vector (i.e., relative position vector) between two points on a circle, then the equation only applies infinitecimally. If Δs is the arc length between two points on the circle, then it works for all values of Δθ. Did you mean to imply that Δs is supposed to be a relative position vector?

FieldvForce, if Chestermiller is correct then you should be aware that a position vector is a path where dr≠0, so you cannot use your incomplete formula. You would need to use the full formula instead. In that sense, it would be inaccurate since you are using a simplified formula in a circumstance which violates the assumption on which the simplification is based.

Thank you very much for clearing things up.

 

[Dale]

You are welcome!