How Can I Define the Bank Angle of a Roller Coaster Helix Track?

[Beyond Aphelion]

Homework Statement

I am working on an overly simplified roller-coaster design but I have found a problem in defining a banked helix track piece.

Essentially I want to be able to track the lateral, normal, and tangential g-forces acting on some point mass, so I need to obtain equations for the tangent and normal accelerations at any given point on the helix.

Homework Equations

It would follow that a helix defined by the curve:

$$\overline{r}(t) = acos(t)\hat{i} + asin(t)\hat{j} + bt\hat{k}$$

$$\overline{T} = \frac{d\overline{r}}{ds} = \frac{d\overline{r}/dt}{ds/dt} = \frac{V}{|V|}$$

$$= \frac{1}{\sqrt{a^{2} + b^{2}}}\left[-asin(t)\hat{i} + acos(t)\hat{j} + b\hat{k}\right]$$

$$\overline{N} = \frac{d\overline{T}/dt}{|d\overline{T}/dt|} = -cos(t)\hate{i} - sin(t)\hat{j}$$

- has a defined principle unit normal vector. However, what if I want to be able to define how steeply the helix is banked ? Essentially, I want to rotate the normal and binormal principle unit vectors so as to reduce as much lateral g-force as possible or eliminate it entirely.

The Attempt at a Solution

My first inclination is that I should introduce a k-hat component (constant if I want the bank angle to be constant or otherwise if not). But does throwing that element into the equation invalidate some other assumptions I have made?

Not knowing the complications that may arise from this problem, I'd like to just define a constant bank angle, but if it is doable, I wonder if changing the bank angle as a function of time/distance would complicate this beyond my level of comprehension.

Another assumption would be that the unit tangent vector would remain the same for any given helix no matter how the bank is defined. However, as the bank is offset, components of acceleration will be transferred between a_t and a_n. Would this even transfer acceleration into my new binormal direction? How would this effect the radius of curvature? I'm just not sure how to go about modeling that.

I don't necessarily have to end up with a concrete equation, but it would be nice. I have MATLAB at my disposal. Any push in the right direction would be helpful.

Thanks.

 

[Beyond Aphelion]

Let's say I keep $$\overline{T}$$ the same.

If I simply redefine $$\overline{N}$$ as $$\overline{N} '$$:

$$\overline{N} ' = -cos(t)\hate{i} - sin(t)\hat{j} - p\hat{k}$$

where 'p' is some constant which defines how steep the track is banked inward, then:

The new unit normal may be obtained from:

$$\frac{\overline{N} '}{|\overline{N} '|} = \frac{1}{\sqrt{1 + p^{2}}}\left[-cos(t)\hat{i} - sin(t)\hat{j} - p\hat{k}\right]$$Now I should theoretically have the binormal unit vector:

$$\overline{B} = \overline{T} x \overline{N} '$$$$=\left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{-asin(t)}{\sqrt{a^{2} + b^{2}}} & \frac{acos(t)}{\sqrt{a^{2} + b^{2}}} & \frac{b}{\sqrt{a^{2} + b^{2}}} \\ \frac{-cos(t)}{\sqrt{1 + p^{2}}} & \frac{-sin(t)}{\sqrt{1 + p^{2}}} & \frac{-p}{\sqrt{1 + p^{2}}} \end{array} \right|$$

Thus,

$$\overline{B} = \frac{1}{\sqrt{(1 + p^{2})(a^{2} + b^{2})}}\left\{\left[-apcos(t) + bsin(t)\right]\hat{i} + \left[-apsin(t) - bcos(t)\right]\hat{j} + \left[a\right]\hat{k}\right\}$$

Does my logic make sense so far?

 

[Beyond Aphelion]

Now to define the radius of curvature, I must find the curvature - which depends on $$\overline{T}$$ parameterized in terms of arc length:

$$\rho = \frac{1}{\kappa}$$

AND

$$\kappa = \left|\frac{d\overline{T}}{ds}\right|$$

In order to find $$\overline{T}$$ in terms of arc length:

$$s(t) = \int_{t_{0}}^{t}\sqrt{[x'(\tau)]^{2} + [y'(\tau)]^{2} + [z'(\tau)]^{2}}d\tau$$

$$= \int_{0}^{t}\sqrt{a^{2}sin^{2}(t) + a^{2}cos^{2}(t) + b^{2}}d\tau = (\sqrt{a^{2} + b^{2}})t$$

So,

$$t = \frac{s}{\sqrt{a^{2} + b^{2}}}$$

may be used as a substitution in $$\overline{T}(t)$$...$$\overline{T} = \frac{1}{\sqrt{a^{2} + b^{2}}}\left[-asin(t)\hat{i} + acos(t)\hat{j} + b\hat{k}\right] \Rightarrow \frac{1}{\sqrt{a^{2} + b^{2}}}\left[-asin(\frac{s}{\sqrt{a^{2} + b^{2}}})\hat{i} + acos(\frac{s}{\sqrt{a^{2} + b^{2}}})\hat{j} + b\hat{k}\right]$$Now we may obtain $$\left|\frac{d\overline{T}}{ds}\right|$$.

$$\frac{d\overline{T}}{ds} = \left[\frac{-a}{a^{2} + b^{2}}cos(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{i} + \left[\frac{-a}{a^{2} + b^{2}}sin(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{j} + 0 \hat{k}$$

$$\kappa = \left|\frac{d\overline{T}}{ds}\right| = \frac{a}{a^{2} + b^{2}}$$So is it right to assume this same curvature for a general helix, even if I am changing the Normal and Binormal unit vectors? Does it truly have no dependence on anything but the change in tangential direction?

 

[Beyond Aphelion]

Torsion, on the other hand, is surely dependent on the Binormal and Normal unit vectors...

$$\tau = -\frac{d\overline{B}}{ds}\bullet\overline{N}$$

Making the substitution $$t = \frac{s}{\sqrt{a^{2} + b^{2}}}$$ again for $$\overline{B}$$ and $$\overline{N} '$$:

$$\overline{B} = \frac{1}{\sqrt{(1 + p^{2})(a^{2} + b^{2})}}\left\{\left[-apcos(\frac{s}{\sqrt{a^{2} + b^{2}}}) + bsin(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{i} + \left[-apsin(\frac{s}{\sqrt{a^{2} + b^{2}}}) - bcos(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{j} + \left[a\right]\hat{k}\right\}$$

$$\overline{N} ' = -cos(\frac{s}{\sqrt{a^{2} + b^{2}}})\hat{i} - sin(\frac{s}{\sqrt{a^{2} + b^{2}}})\hat{j} - p\hat{k}$$I can see how this might get ugly if I vary 'p' as a function of time/arc-length...

$$\frac{d\overline{B}}{ds} = \frac{1}{(a^{2} + b^{2})\sqrt{(1 + p^{2})}}\left\{\left[apsin(\frac{s}{\sqrt{a^{2} + b^{2}}}) + bcos(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{i} + \left[-apcos(\frac{s}{\sqrt{a^{2} + b^{2}}}) + bsin(\frac{s}{\sqrt{a^{2} + b^{2}}})\right]\hat{j}\right\}$$

To make things easier to see:

Let: $$\frac{1}{(a^{2} + b^{2})\sqrt{(1 + p^{2})}} = \omega$$

AND

Let: $$\frac{s}{\sqrt{a^{2} + b^{2}}} = \varphi$$

Then,

$$\tau' = -\frac{d\overline{B}}{ds}\bullet\overline{N} ' = -\omega\left\{\left[apsin(\varphi) + bcos(\varphi)\right]\hat{i} + \left[-apcos(\varphi) + bsin(\varphi)\right]\hat{j}\right\}\bullet\left\{-cos(\varphi)\hat{i} - sin(\varphi)\hat{j} - p\hat{k}\right\}$$

$$=\omega\left[apsin(\varphi)cos(\varphi) + bcos^{2}(\varphi) - apsin(\varphi)cos(\varphi) + bsin^{2}(\varphi)\right]$$

So,

$$\tau' = \omega b = \frac{b}{(a^{2} + b^{2})\sqrt{(1 + p^{2})}}$$To check the validity of this torsion value, I can set 'p = 0' as if I had not banked the helix:

$$\tau' = \frac{b}{a^{2} + b^{2}}$$

This is what I was expecting (for a general helix without banking). So I am at least a little more confident in my approach. But I am not done yet...

 

[Beyond Aphelion]

Any ideas on where I should go from here for my g-force analysis?

 

[Beyond Aphelion]

I just realized my entire method is wrong because I didn't take into account the fact that:

$$v(z) = \sqrt{2g(z_{0} - z)}$$ (at all times on the rollercoaster)

Where $$z_{0}$$ is the height at which the frictionless rollercoaster started.

So now my approach has altered slightly:

$$\overline{r}(\theta) = Rcos(\theta)\hat{i} + Rsin(\theta)\hat{j} + z\hat{k}$$

In order to use my velocity equation derived from the concept of energy, I must get my position equation in terms of z. I changed the position equation in terms of theta to avoid dealing with time for now as that introduces some confusion.

Since $$\frac{dz}{d\theta} = constant = \frac{M}{2\pi}$$,

where M is the $$\frac{\Delta z}{revolution}$$ of the helix. We can define M however we want depending on the shape of the helix.

$$\int_{0}^{z'}dz = \int_{0}^{\theta '}\frac{M}{2\pi}d \theta \Rightarrow z = \frac{M}{2\pi}\theta \Rightarrow \theta = \frac{2\pi z}{M}$$


$$\overline{r}(z) = Rcos(\frac{2\pi z}{M})\hat{i} + Rsin(\frac{2\pi z}{M})\hat{j} + z\hat{k}$$


Now how do I go about getting z in terms of time so that I can get position in terms of time for the helix?