Velocity of wind based on the shape of a smoke trail

[weirdoguy]

Hi everyone. My high-school student got the following homework exercise at school which I have problem with:

Smoke from the steam locomotive is carried by horizontally blowing wind. Shape of the smoke trail is shown in the attached figure. Using the drawing, determine velocity of the wind assuming it is constant. Speed of the steam engine is $v=60\frac{km}{h}$ and it is moving on an arc of a circle.

My problem is that I really don't see any way to approach it on an elementary level. If there were no wind the smoke trail would be an involute of circle. I derived parametric equation of the trail for radius $R=1m$ and speed of the train $v=1\frac{m}{s}$ with $\vec{v}_{initial}=[0,1]$, $\vec{r}_{initial}=[1,0]$ and $\vec{v}_{wind}=[v_{xw},v_{yw}]$:
$x(t)=\cos t+(T-t)(v_{xw}-\sin t)$
$y(t)=\sin t +(T-t)(v_{yw}+\cos t)$
for $t\in\langle 0,T\rangle$.

Playing with these a little bit I've noticed that the cusp seen in the attached picture appears only for quite specific range of values of $v_{xw}$ and $v_{yw}$. For others there is a loop, and sometimes none of these. Anyway, it's still not clear for me how to approach this problem on a high-school level

Any thoughts?

 

[A.T.]

If there were no wind the smoke trail would be an involute of circle.

Why not just a circle?

Speed of the steam engine is v=60kmh and it is moving on an arc of a circle.

That's all that is given?

Anyway, it's still not clear for me how to approach this problem on a high-school level

Two observations that might lead to a solution, without complex math:
- The trail leaves the locomotive perpendicular to the track (not sure if meant that way)
- The trail has a sharp corner

The sharp corner could imply that the wind is also 60km/h, because there is a point along the track with no relative wind at the locomotive.

 

[BvU]

The sharp corner is an indication that train and wind had the same speed at some point on the arc. That point was where wind direction was tangent to the train trajectory. Both speeds are constant, so wind speed = train speed.

 

[A.T.]

The sharp corner is an indication that train and wind had the same speed at some point on the arc. That point was where wind direction was tangent to the train trajectory. Both speeds are constant, so wind speed = train speed.

Exactly. And I think the shape of the trail is a cycloid:
https://en.wikipedia.org/wiki/Cycloid

The shape of the trail is the trajectory of the locomotive in the rest frame of the airmass:

 

[Dale]

I've noticed that the cusp seen in the attached picture appears only for quite specific range of values of vxw and vyw.

Interesting. Is it a range of values or is it just 60 km/h?

 

[weirdoguy]

Why not just a circle?

Well, I assumed that smoke released at a point on the arc moves with constant speed (equal to speed of the train) along the tangent but now I see that this was bad assumption Thanks guys!

Funny enough, I did quite similar exercise with boats on a river not so long ago and I didn't make that mistake.

 

[weirdoguy]

Interesting. Is it a range of values or is it just 60 km/h?

Since my assumptions regarding behaviour of smoke were wrong, I guess it doesn't matter

 

[A.T.]

Funny enough, I did quite similar exercise with boats on a river not so long ago and I didn't make that mistake.

Without wind, a steam boat on a river, that goes in circles at the same speed as the river flows, will produce the same smoke trail.

 

[etotheipi]

You already know this by now but I was just going to say that I'd expect the smoke to almost immediately acquire the velocity of the wind, i.e. the smoke has zero velocity in the rest frame of the air mass once it has left the train. So the trail left parameterised by $t$ will be$$\vec{x} = [\cos{t} + (T-t)v_{xw},\sin{t} + (T-t) v_{yw}]$$which looks like a rotated cycloid.

 

[A.T.]

Shape of the smoke trail is shown in the attached figure.

It's a nice puzzle. The only small quibble I have with the figure: The thick dashed line might suggest that it represents smoke clouds released at constant frequency. Under that interpretation, the uniform spacing and size of the dashes would be misleading.

 

[weirdoguy]

Without wind, a steam boat on a river, that goes in circles

There were actually two boats traveling along a canal that is joining two lakes (sorry, I rarely use words "river" and "lake" and I mix their meaning) - see attached picture.

Ratio of their speeds is $\frac{v_1}{v_2}=\frac{3}{5}$ and one has to determine direction of the wind. The only "problem" with this exercise is that no one said that this drawing shows situation from an angle not perpendicularly from above and author assumes that wind blows parallel to canal. But besides that, it's quite simple and I didn't have any problem with how smoke behaves. I mean, I've seen smoke couple of times I don't know why I assumed different behaviour in that exercise with train. Two days ago I was thinking about this problem so intensely that I had problem falling asleep and when I did i had a dream in which it appeared

 

[A.T.]

There were actually two boats traveling along a canal that is joining two lakes (sorry, I rarely use words "river" and "lake" and I mix their meaning) - see attached picture.
View attachment 274586
Ratio of their speeds is $\frac{v_1}{v_2}=\frac{3}{5}$ and one has to determine direction of the wind. The only "problem" with this exercise is that no one said that this drawing shows situation from an angle not perpendicularly from above and author assumes that wind blows parallel to canal. But besides that, it's quite simple and I didn't have any problem with how smoke behaves. I mean, I've seen smoke couple of times I don't know why I assumed different behaviour in that exercise with train. Two days ago I was thinking about this problem so intensely that I had problem falling asleep and when I did i had a dream in which it appeared

Can you post the problem statement? It's OK if it is not English.

 

[weirdoguy]

Figure 4, shown from a bird's eye view, shows traces of smoke from the chimneys of two ships, which are sailing in opposite directions through the canal connecting two lakes. Determine the wind direction if the ratio of their speeds is: $\frac{v_1}{v_2}=\frac{3}{5}$.

I've actually searched "bird's eye view" in google and some pictures are taken from an angle, and some are perpendicular to the surface of Earth, so it's not necessarily obvious for me what author meant.

 

[A.T.]

Figure 4, shown from a bird's eye view, shows traces of smoke from the chimneys of two ships, which are sailing in opposite directions through the canal connecting two lakes. Determine the wind direction if the ratio of their speeds is: $\frac{v_1}{v_2}=\frac{3}{5}$.

Not sure what the canal connecting two lakes means to imply. But assuming there is no flow in the canal, the wind direction is along the line from their meeting point (3/8 of the way from boat 1 to boat 2) and the crossing of the smoke trails.

 

[weirdoguy]

Not sure what the canal connecting two lakes means to imply.

Based on the solution provided by the author - nothing.

 

[Steve4Physics]

It's a nice puzzle. The only small quibble I have with the figure:
...

You can't beat a good quibble. So I'll also add that the 'height' of a cycloid should equal the diameter of the generating circle. The portion of the cycloid in the diagram is only about half as 'high' as it should be.

Or the circular track is twice as big as it should be.

Depending on if you're a cycloid-half-full or cycloid-half-empty kind of a person.

Sorry, I'm rambling. Season's Greetings.

 

[BvU]

The portion of the cycloid in the diagram is only about half as 'high' as it should be.

Is it really that bad ?