SAT Question Everyone Got Wrong

[Albert Chan]

[Maximilian Hohmann]

Hello!

Excellent as always. Thanks for posting this.

Regards
Max

[teenix]

This next is definitely off topic, and this thread I suppose, but the thread title popped a memory in my head :-)

When I was a flight instructor, I used to pose a question to students (and some aircraft mechanics) related to the operation of the piston engine they were flying with.

These engines are fitted with a manually controlled fuel mixture lever which is used to lean the fuel/air ratio depending on the altitude you are flying at. The higher you go, the less air pressure available, so the engine will start to run rich unless you compensate.

Leaning a fuel mixture also increases the burn temperature, but only up to a peak, then the temperature starts dropping again - less fuel = less burn. Usually, but not always, an engine is set to run just before the peak temperature occurs. You use the exhaust gas temperature gauge to set this.

The problem that students get taught is that if you run an engine on the lean side of the peak, engine damage can occur because aluminum parts like pistons and perhaps valves and or their seats can burn and may lead to horrible consequences.

So I posed the question.

If you have the fuel mixture leaned to the hottest temperature possible, and continue to lean it while the engine and exhaust temperatures both begin reducing until the engine finally stalls, why does it cause an overheating problem? Those in the know should come up with the answer, but no-one I asked ever did.

PS
You got to love the English language, at school you are taught I before E except after C, so why is "thier" spelt wrong?

cheers

Tony

[Maximilian Hohmann]

Hello Tony,

(11-30-2023 06:27 PM)teenix Wrote:  So I posed the question.

If you have the fuel mixture leaned to the hottest temperature possible, and continue to lean it while the engine and exhaust temperatures both begin reducing until the engine finally stalls, why does it cause an overheating problem? Those in the know should come up with the answer, but no-one I asked ever did.

Do you already know the answer or do you want to hear it?

Regards
Max

[teenix]

(11-30-2023 07:46 PM)Maximilian Hohmann Wrote:  Do you already know the answer or do you want to hear it?
Regards
Max

Hi Max,

Yes, I figured it out. I thought it was a contradictory situation and might promote some head scratching.

[KeithB]

(11-30-2023 06:27 PM)teenix Wrote:  You got to love the English language, at school you are taught I before E except after C, so why is "thier" spelt wrong?

cheers

Tony

There are a lot of exceptions to that rule
Keith

[Paul Dale]

A beautiful example of epicyclic gearing which is really useful to get difficult ratios.
Getting a ratio of 1 : (a+b) is immensely helpful when you are targeting specific ratios.

E.g. a lunar phase indicator in a clock or an orrery. Here's one of the latter I prepared earlier: http://www.nzmeccano.com/image-83067

[jte]

I didn’t get a chance to watch the video yet, but my way of looking at this is to first have a rotating reference frame, as if the two circles are on fixed axles (keeping the centers of the circles and the point of contact all at fixed positions). It’s then intuitive that the ratio of circumferences determines the number of times the smaller circle rotates (to the one turn of the larger). Then one can rotate the reference frame to get the one additional turn.

[teenix]

I have never searched for anything on UTube from the presenter of the SAT video, but I was looking at UTube this afternoon, and out of the blue there was the same video listed, which is somewhat weird. I failed though, I would have chosen the wrong answer - 3.

If anyone is interested in the engine problem, the answer is fairly simple. Even though the engine and exhaust gas temperatures fall when the mixture is leaned past the peak, any fuel that is still burning is quite hot. Even though it may be hot, the burn is in a smaller area, so the rest of the engine and the exhaust gasses are cooler than normal - less energy overall.

However, if that hot burn is consistently in the one area in the combustion chamber, it is possible that that area gets too hot and can melt metal, ie a hole in the piston. Pre-ignition (pinging) may also occur because of the hot spot which can cause mechanical damage as well. Easy to hear in a car, but not so in a noisy cockpit.

cheers

Tony

[Albert Chan]

Hi, teenix

If we flatten the big circle as a straight line, small circle indeed travelled 3 times its perimeter.

  ◒__________◒__________◒__________◒

But, if we bent the track with its end-to-end touched, we get an extra revolution.
We can bent it to any shape, not just a big circle.

  ◒
  ┌──────────┐◐
  │          │
  │          │
  │      ↻    │
  │          │
  │          │
  ◑└──────────┘
            ◓

[klesl]

translation movement only - 3 revolutions
CW translation and CW rotation movements - 4 revolutions
But what's happened if the translation movement is CW and the rotation movement is CCW (backward)?

[Albert Chan]

(12-03-2023 12:58 PM)klesl Wrote:  translation movement only - 3 revolutions
CW translation and CW rotation movements - 4 revolutions
But what's happened if the translation movement is CW and the rotation movement is CCW (backward)?

If small circle is rotating inside a region, its rotation is CCW (backward)
But small circle gain a CW rotation. (black part always touch the "floor")
In other words, we lost 1 CCW revolution.

  ┌──────────┐
  │◓        ◑│
  │          │
  │      ↻    │
  │          │
  │◐        ◒│
  └──────────┘            

[Maximilian Hohmann]

Hello!

(12-03-2023 11:11 AM)teenix Wrote:  I have never searched for anything on UTube from the presenter of the SAT video, ...

You are missing a lot! Derek Muller, an Australian, got his ph.d. in physics for a thesis about effective multimedia for physics education (https://citeseerx.ist.psu.edu/document?r...ed4d9a4209) and it shows in all of his videos. There must be very few of his videos that I have not seen yet.

Absolutely outstanding is his 3-hour-long (in three parts) documentation about Uranium (https://en.wikipedia.org/wiki/Uranium_%E...n%27s_Tail) which was shown on German National television some years ago. This is when I learned about him and started watching his youtube videos.

Or this one about the single person who killed the most people by playing a major part in inventing both leaded fuel and chlorofluorcarbons: https://www.youtube.com/watch?v=IV3dnLzthDA
Interesting for both of us because, if I understand it correctly, we both work in one of the few sectors where leaded fuel still exists.

Regards
Max

[teenix]

Indeed, afterwards I also watched The Bizarre Behavior of Rotating Bodies by the same presenter. I found that quite interesting.

https://www.youtube.com/watch?v=1VPfZ_Xz...luZw%3D%3D

cheers

Tony

[ijabbott]

(12-03-2023 03:05 PM)Albert Chan Wrote:  

(12-03-2023 12:58 PM)klesl Wrote:  translation movement only - 3 revolutions
CW translation and CW rotation movements - 4 revolutions
But what's happened if the translation movement is CW and the rotation movement is CCW (backward)?

If small circle is rotating inside a region, its rotation is CCW (backward)
But small circle gain a CW rotation. (black part always touch the "floor")
In other words, we lost 1 CCW revolution.

  ┌──────────┐
  │◓        ◑│
  │          │
  │      ↻    │
  │          │
  │◐        ◒│
  └──────────┘            

And for a circle rotating inside any shape with corners, one cannot merely subtract 1 from the quotient of the perimeter of the shape and the circumference of the circle, because the interface between the shapes is no longer continuous. Instead, for a rectangle you would need to subtract four circle radii from the perimeter of the rectangle, and divide that by the circumference of the circle.

[Albert Chan]

(12-03-2023 10:38 PM)ijabbott Wrote:  And for a circle rotating inside any shape with corners, one cannot merely subtract 1 from the quotient of the perimeter of the shape and the circumference of the circle, because the interface between the shapes is no longer continuous. Instead, for a rectangle you would need to subtract four circle radii from the perimeter of the rectangle, and divide that by the circumference of the circle.

You are right! There is a limit of bending of the tracks.

I assumed small circle does not simultaneously touch 2 "floors".
If it does, then we have to analyze the trip piecewise, to remove effect of skipped corners.
Note: this applied to small circle rotating outside irregular shape with valleys as well.

If small circle is rotating inside a square (or rectangle) with 3 times circle perimeter:
tan(90°/2) = (d/2) / (x/2) --> x = d

Revolutions = 4 * (3*pi*d/4 - d) / (pi*d) = 3 - 4/pi ≈ 1.727

If small circle is rotating inside an equilateral triangle with 3 times circle perimeter:
tan(60°/2) = (d/2) / (x/2) --> x = √3*d

Revolutions = 3 * (3*pi*d/3 - √3*d) / (pi*d) = 3 - 3*√3/pi ≈ 1.346

Without considering skipped corners, we would get (for both cases), revolutions = 3 - 1 = 2

[Jonathan Busby]

I think the confusion that arises over this question is due to the accidental and almost subconscious conflation of the rotational behavior of a disc on a perfectly flat, level surface with the rotational behavior of a disc over a *curved* surface.

When a disc rotates over any surface, the number of rotations is equal to the distance traced out by the center of the disc divided by the circumference of the disc. Now, on a perfectly flat surface, the distance traced out by the center of the disc is equal to the distance the disc has traveled. So, in this case, one can divide the distance traveled by the disc by its circumference to arrive at the number of rotations. So, on a curved surface, such as another disc in this case, one must be able to find the number of rotations by dividing the stationary disc's circumference by the circumference of a moving disc, right? Right? Nope The key mistake in this reasoning is ignoring the distance the center of the peripheral disc has to traverse to make a full circuit around the stationary disc. Since the number of rotations the peripheral disc makes is equal to the distanced traversed by its center divided by its circumference, we arrive at the following :

Size of circle traversed by center of moving disc :

\[ \displaystyle (r_{m} + r_{s}) \cdot 2 \cdot \pi \]

Where \( \displaystyle r_{m} \) is radius of the moving disc and \( \displaystyle r_{s} \) is the radius of the stationary disc.

Now, since the number of rotations the disc makes is equal to the distance traveled by its center divided by its circumference, we have :

\[ \displaystyle \dfrac{((r_{m} + r_{s}) \cdot 2 \cdot \pi )}{r_{m} \cdot 2 \cdot \pi} = \]


\[ \displaystyle 1 + \dfrac{r_{s}}{r_{m}} \]


Hope that clears it up for somebody

Regards,

Jonathan

[Werner]

There's a similar one (in that almost everyone, including the test designers, got it wrong):



This time though, the correct answer did appear among the choices.

Cheers, Werner

[Albert Chan]

Paper models to count faces when joining tetrahedron and pyramid, is not convincing enough.
We could have what is claimed as 1 face really is 2, just visually too close to call.

From https://math.stackexchange.com/a/1856243, adding an extra pyramid is much better.
Pink bottom angle = 180° - 2*60° = 60°, which make Pink region a tetrahedron.

[BruceH]

(12-05-2023 04:34 PM)Werner Wrote:  There's a similar one (in that almost everyone, including the test designers, got it wrong):

Another way of looking at those diagrams in test papers. :-)
https://xkcd.com/2509/