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Challenge Questions

Stop in and exercise your brain. Talk about this month's Challenge from Specs & Techs or similar puzzles.

So do you have a Challenge Question that could stump the community? Then submit the question with the "correct" answer and we'll post it. If it's really good, we may even roll it up to Specs & Techs. You'll be famous!

Answers to Challenge Questions appear by the last Tuesday of the month.

Apollo 11 Communications Log- Gimbal Lock: Newsletter Challenge (July 2015)

Posted July 01, 2015 12:00 AM
Pathfinder Tags: challenge questions NASA

This month's Challenge Question: Specs & Techs from IHS Engineering360:

CapCom - "Columbia, Houston. We noticed you are maneuvering very close to gimbal lock. We suggest you move back away. Over." Michael Collins (CMP) - "Yes. I am going around it...How about sending me a fourth gimbal for Christmas?"

Why did Michael Collins ask for a fourth gimbal for Christmas?

And the answer is:

The full communication log can be found here: http://apollo11.spacelog.org/page/04:08:40:46/
In the discussion above, mission control (CapCom) is communicating to Michael Collins who is piloting the Apollo Command/Service Module, that he is nearing gimbal lock. This is in reference to the Inertial Measurement Unit (IMU) which has Outer, Middle, and Inner gimbals (see diagram below). The outer gimbal is mounted on the navigation base which in turn is rigidly mounted to the spacecraft.

Michael Collins had flown in Gemini 10, which had a fourth or "redundant" gimbal. This redundant gimbal was mounted outside the normal outer gimbal. Normal operation would then be to use the inner three gimbals to drive the stabilizing gyro error signals to zero while the fourth gimbal could be used to keep the middle gimbal near zero and away from the gimbal lock orientation.

The general lesson is, rotations in 3D in space are tricky. That is why 3D simulators are usually based on quaternions rather than Euler Angles. Quaternions have a built-in extra degree of freedom that avoids gimbal lock. Essentially adding a fourth (redundant) gimbal does the same thing.

11 comments; last comment on 07/21/2015
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Airbag Physics: Newsletter Challenge (June 2015)

Posted June 01, 2015 12:00 AM
Pathfinder Tags: challenge question

This month's Challenge Question: Specs & Techs from IHS Engineering360:

This month we had the largest automotive recall in U.S. history: 33.8 million vehicles made by 11 different automakers were recalled due to malfunctioning frontal airbags. These airbags were made by the parts supplier Takata. So, let's think about airbags for this month's challenge question.

We know that airbags are very soft inflatable devices; during a crash the airbag is the best accessory to save a life. The question this month is: Why? Why do airbags save lives? It is not enough to say that lives are saved during a crash because the airbags are soft devices. I am looking here at the physics that supports this notion.

And the answer is:

Let's assume, for instance, that a car is moving at 60 mph and crashes into a tree. A passenger in the front seat, before the crash, is moving at 60 mph. At the time of the crash their body will stop (normally by hitting the dash of the car, if there is no airbag). So the passenger momentum will change from mv to 0, where v = 60 mph.

Now, in order to change momentum, according to Newton, a force needs to act for a given amount of time. This concept is a consequence of Newton's second law, which is stated as follows

F = ma

where a is the acceleration of an object of mass m and F is the force generated by this acceleration. We know that acceleration is a change in velocity/time, so we can write the above expression as follows

and finally,

The expression on the left is called the impulse. So the impulse is equal to the change in momentum. t is the time we must apply the force F in order to produce a change in momentum.

The equation can be written as,

In the case of the crash in a car without airbags, at the moment of the crash the passenger will start moving at 60 mph toward the dash. At the very moment that the passenger hits the dash, a force F will be applied by the dash to the passenger (Newton's first law) and in practically zero time (because the dash does not subside) the speed of the passenger goes to zero. Looking at the last equation, we see that the force the dash is applying to the passenger is almost infinity. This, of course, will kill the passenger.

If airbags are available, the passenger will hit the airbag instead of the dash. As soon as the passenger hits the airbag, a force F will be applied to the passenger by the bag, but because the airbag is flexible this force will be applied during a finite time (not zero). This is the time measured from the moment the passenger touches the bag until the speed of the passenger is zero. Because the time is not zero, we see from the above equation that the force is not infinity.

Therefore airbags save lives because they allow the force used to stop the speed of the passenger to act for a certain period of time.

27 comments; last comment on 06/06/2015
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Styrofoam Slide (May 2015)

Posted May 01, 2015 12:00 AM
Pathfinder Tags: challenge question

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Try this experiment: slide a Styrofoam cup full of water across a smooth (finished) wood surface. Make sure the speed is around 10 cm/s. You will notice that at this speed droplets of water from the cup shoot up to about 20 cm. Explain why.

And the answer is:

As you slide the cup at this speed, energy couples into the sliding cup. The almost-flat bottom of the cup slips and sticks to the surface, producing nonlinear vibrations. This is exactly a relaxation oscillator (producing a non-sinusoidal repetitive signal) that immediately produces standing waves in the surface of water. As you continue pushing the cup the water crests of the standing wave increase in height, causing drops of water to break from the surface and be projected into the air.

49 comments; last comment on 06/27/2015
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100 Meter Dash: Newsletter Challenge (April 2015)

Posted April 01, 2015 12:00 AM
Pathfinder Tags: challenge question

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Now that good weather is around the corner, a challenge question related to outdoor activities is a must. In short races (100 meters or less) breathing is not necessary for the runner. Why?

And the answer is:

Runners (and all of us, as well) need chemical energy in their muscles. This energy is made available by aerobic (with oxygen) and anaerobic (without oxygen) reactions. However, after breathing, oxygen is converter to chemical energy in the muscles in a period of time that is longer than the time it takes to finish a short race (ten or less seconds). The oxygen inhaled before the start of the race is sufficient to provide the aerobic energy needed for the race.

38 comments; last comment on 04/28/2015
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3464-Cube: Newsletter Challenge (March 2015)

Posted February 28, 2015 5:01 PM

This month's Challenge Question: Specs & Techs from IHS Engineering360:

A 2-cube (square) has 4 vertices and 4 edges. A 3-cube (cube) has 8 vertices, 12 edges, and 6 faces. A 4-cube (tesseract) has 16 vertices, 32 edges, 24 faces, and 8 cells. In three dimensions, the Euler characteristic says that vertices + faces - edges = 2. What is the right side of the Euler characteristic equation equal to when generalized and calculated for a 3464-cube?

And the answer is:

The answer is 0. The vertex (0-face), edge (1-face), face (2-face), cell (3-face), 4-face, etc. are the smaller dimensional objects that make up a larger dimensional shape. There are always n-1 dimensional objects that make up an n dimensional shape.

For instance, in 2 dimensions, a square is made up of 1 dimensional lines (edges) and the zero dimensional points (vertices). A three dimensional cube is made up of 2 dimensional sides (faces), 1 dimensional lines (edges) and zero dimensional points (vertices). When the Euler Characteristic is generalized for all dimensions, it says add up the number of even dimensional components and subtract from them the number of odd dimensional components. For an n-cube, that difference will be 0 if n is even and 2 if n is odd. So in the case of a cube in three dimensions, vertices + faces - edges = 2. For a 3464-cube, vertices + faces + 4-faces +…+ 3462-faces - edges -cells - 5-faces - … - 3463-faces = 0.

http://en.wikipedia.org/wiki/Hypercube#Elements (See Hypercube Elements Chart)

Editor's Note: This question was previously titled "3463-Cube." It was an error and has since been corrected to "3464-Cube."

21 comments; last comment on 03/17/2015
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Jovian System: Newsletter Challenge (February 2015)

Posted February 01, 2015 12:00 AM
Pathfinder Tags: challenge question

This month's Challenge Question: Specs & Techs from IHS Engineering360:

It's 2045 and a group of space pioneers are on an interplanetary transport to the Callisto colony. The pioneers were awoken from a four-year hibernation, having traveled to Jupiter's orbit only to find a large dumbbell-shaped asteroid several hundred kilometers wide where they expected Jupiter to be. Where are they?

And the answer is:

The transport must have had a critical navigation error early in the trip because it sounds like they have reached Jupiter's leading Lagrangian point, L4 , and are viewing 624 Hektor, the largest of Jupiter's Trojans.

"624Hektor-LB1-mag15" by I, Kevin Heider. Licensed under CC BY-SA 3.0 via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:624Hektor-LB1-mag15.jpg#mediaviewer/File:624Hektor-LB1-mag15.jpg

39 comments; last comment on 02/18/2015
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