The dumb questions thread

[steve_bank]

In rfeliabilty there is the 'bath tub curves'. Originally derived from hiuman death statistics is applied to systems where it fits.

In the bath tub at birth thru say 8-10 there is an initially high mortality rate that drops exponentially. In electronics we use term infant mortality for early failures. Early deaths are disease, birth defects and so on. As people get older the curve falens out to a normal distribution, called chance failure. Hit by a car, catch a disease from someone and so on. As one gets older the mortality rate increases exponentially and in electronics we call that wear out. Things begin to break down. Heart dsieas and so om.

Complex electronic systems tend to follow the bah tub curves. It is the Weibull destruction. The idea that combinations of components of different failure distribution average out to n apparent random phenomena. Individual components' do not necessarily have normal distributions. I have seen the bath tub curve from reliability on systems.

Something like a battery may have an increased probability of failure based on usage. You may find failure curves on line somewhere.

The failure rate of system is 1/f1 + 1/ f2..= 1/ft where f are failure rates. Assuming a bath tub curve and no infant failures. It is how the failure rate of a circuit board with many parts are calculated.

You might find an old reliability standard Mil-217 online that goes through the basic theory. Other failure distributions are log normal and exponential for reliability.

https://www.sciencedirect.com/topics/engineering/bathtub-curve

 

[Bomb#20]

So here's the question: What are the odds that all three batteries will die at precisely the same time down to only a second or two difference?

One in a trillion, near enough.

Can you explain for a layperson why that is?

Sure. (Steve and bilby are vastly underestimating the odds because they aren't answering the right question. Bilby is calculating the chance of two batteries dying at the same time, not three. Steve isn't looking at independent events -- the airplane backup generator failed at the same exact time as the main generator because until the main generator failed the backup wasn't being called on to do any work. It could have been nonfunctional for hours or days beforehand and nobody noticed because it wasn't powering the plane.)

As far as the actual calculation goes, bilby's approach is correct, but his numbers are, as he said, ex ano. A normal 9-volt battery holds about 500 mAh; in smoke detectors they last a lot longer than 6 months. (Manufacturers recommend a 6-month replacement cycle to cover their asses, not because it's necessary.) He's guessing about a half a percent for the standard deviation in battery life; I have no idea whether battery manufacturing is really that precise but I know electronics manufacturing isn't, so if he's right that batteries are that consistent then that just means the lifetime variation will be dominated by the variation in current-draw of the smoke detectors. For normal electronics the standard deviation will be something like 5 percent. So he's estimating the difference in lifespans to be typically hours; weeks is more realistic. That's on the order of a million seconds, so the chance of two dying within a second of each other is on the order of one in a million. That's for two batteries; to get the answer for three you have to square that, hence, one in a trillion.

 

[steve_bank]

A common mistake among non practitioners is to look at probabilities subjectively.

The numbers are what they are. Flip 2 coins a number of times, what is odds of at least one coming up heads?

Carrying it further flip 100 coins at once. Any 1 head is considered good. No heads is a failure. What are the odds of all 100 coins coming up tails, low but not zero. Run the trials enough and it will happen.

Reliability in general is the same. In the 80s in one of my incarnations I was a reliability engineer in the systems group they used to have. Computers and the old Multibus architecture.

When systems are properly stressed in manufacturing, meaning infant or early failures are eliminated, over time a large group of systems will have a constant failure rate, or the hazard rate. MTBF mean time between failure. The same percentage of systems will fail say every month within statistical variation.

If you send an email to support at a battery company they will give you reliability data on their products. It is not a mystery.

The probability of a battery lasting x hours before failures is x, two redundant batteries is y. Just like the coin example.

An example I have used in the past is incandescent light bulbs. Imagine a wall of light bulbs from manufactures with differnt failure rates and distributions. When a light bulb fails reokace it with a bukb from any source. Regardless of underlying distributions the number of bulb failures say per month will be the same with some variation of course.

There is no way to predict exactly which bulbs will fail and when, but on the average the percent failure per unit time will be constant.

So, when considering a single battery it can fail at any time. Take a large number of batteries and the percent that fail per unit time is predictable. Use two redundant batteries and on the average the reliability will be bettr than a single battery, but both can fail at any time.

An old textbook problem. For a mission there needs to be 20 bombers arriving at a target, Each plane has a failure rate. How many aircraft have to take off to ensure 20 arrive at the target?

Statistics is all about averages.

 

[Angry Floof]

Thank you, Bomb, Steve, and Bill. I knew I could count on you all to answer the question.

 

[fromderinside]

You ask an important question Angry_Floof.

I like to look at the problem form a safety view. We want the plane to safely land so we build a system of controllers. We worry about system failure so we can increase the probability for each and every controller surviving a failure for landing and we additionally add several controller systems to each landing control.

We test controllers for high reliability for long life. And we provide three independent systems for control to minimize external failure while increasing internal reliability further. We could have left things at two systems, but there are possible external events might overlap the two systems at the same time. By adding the third level of redundancy we thought we maximized probability of equipment safety reducing the likelihood that systems will fail to all systems to where the likelihood that an inspection or maintenance action will miss something is reduced to the level where it is very unlikely that all three systems will fail.

Well except for the DC-10 over Iowa which threw an engine vane through the containment vessel into the tail where all three systems came together for servicing of tail direction and attitude surface control components back in ought-89.

The pilot was the final safety factor and through his skill he saved many lives on that crash landing.

In fact by considering this topic of life and failure engineering has become so sophisticated that we now design things to wear into tolerance rather than worrying about keeping it in tolerance.

 

[Angry Floof]

New dumb question. You have a coffee maker turned on with the 12-cup carafe full of hot water. It's the type of coffee machine that has a hot plate under the carafe to keep it hot until it's turned off.

It's in a room that is about 10' x 20', or 3x6 meters.

The dumb question is: will that coffee pot full of hot water sitting on all day increase the temperature of that room? Say the door stays shut most of the time, maybe being opened and closed ten times over the course of the day.

 

[bilby]

New dumb question. You have a coffee maker turned on with the 12-cup carafe full of hot water. It's the type of coffee machine that has a hot plate under the carafe to keep it hot until it's turned off.

It's in a room that is about 10' x 20', or 3x6 meters.

The dumb question is: will that coffee pot full of hot water sitting on all day increase the temperature of that room? Say the door stays shut most of the time, maybe being opened and closed ten times over the course of the day.

Yes, it will.

Probably not by very much.

Whether it's noticeable depends on a lot of factors. In a hypothetical sealed room with no way for the heat to get out, the room will eventually reach the temperature that the coffee pot thermostat is set to. Obviously real rooms are not even close to that scenario; They have air circulation from outside in various ways, and also lose or gain heat via conduction and radiation.

A human being gives off about 100W of energy. The best estimate I can find for a coffee pot keeping warm online is around 30Wh over 8 hours, or about 4W - so if that's right, four coffee jugs would heat the room about as much as an extra person.

So if a coworker is complaining that your coffee jug is making the room too warm, getting them to fuck off is about four times as effective in cooling the room down as turning the pot off.

 

[steve_bank]

The short answer is yes it will just lie an eelctric space heater. How much depends on conditions.,

The long answer.

How much depends on what is called thermal resistance between the room and the what is outside room. The mass of the EARTH is called an ultimate heat sink, large realtive mass and ignoring radiation.

A simplr 1st order model . Imagine a resistor from the hot plate to the air, the air to the wall, across the wall, and the wall to te thermal mass of the Earth.

Electric current through a resistor causes a voltage rise. In heat transfer heat flow in watts replaces current and a temperature rise develops across each thermal resistance. 1 watt of heat flowing through a 1 deg/watt thermal resistor results in a 1 deg temp rise.

If the room is well insulated and the mass of the air is a lot less than the walls the equation q = m*c * dt approximates the rise.

Q = heat
m = mass
c = specific heat of air
dt change in temp

 

[Angry Floof]

So yes but only a tiny, unnoticeable bit due to vents and doors opening from time to time. Thanks, guys!

 

[Loren Pechtel]

New dumb question. You have a coffee maker turned on with the 12-cup carafe full of hot water. It's the type of coffee machine that has a hot plate under the carafe to keep it hot until it's turned off.

It's in a room that is about 10' x 20', or 3x6 meters.

The dumb question is: will that coffee pot full of hot water sitting on all day increase the temperature of that room? Say the door stays shut most of the time, maybe being opened and closed ten times over the course of the day.

Certainly, the only question is by how much.

If you want more information plug the coffeemaker into something that can monitor power usage. (The original example was the Kill-O-Watt, but many such products exist these days.) Note what the power usage is while it's keeping it warm. The effect will be the same as if you were running an electric heater of that power in the room.

 

[steve_bank]

There are two perfectly parallel vertical plates separated by 1 meter. Initial conditions a small ball at 10 meters high is starting away from one plate directly at the other plate.


Initial velocity is 100m/s. Ball mass is .01kg. Assume the system is perfect, the ball always travels in one plane when bouncing.

The plates and ball are hard with negligible loss of kinetic energy for the ball bouncing off the plates. Ignore air resistance.

1. how long before the ball hits the ground.
2. How many bounce between the plates.
3. Where between the plates does the ball hit the ground.

 

[Loren Pechtel]

There are two perfectly parallel vertical plates separated by 1 meter. Initial conditions a small ball at 10 meters high is starting away from one plate directly at the other plate.


Initial velocity is 100m/s. Ball mass is .01kg. Assume the system is perfect, the ball always travels in one plane when bouncing.

The plates and ball are hard with negligible loss of kinetic energy for the ball bouncing off the plates. Ignore air resistance.

1. how long before the ball hits the ground.
2. How many bounce between the plates.
3. Where between the plates does the ball hit the ground.

Ball mass is irrelevant. The ball falls exactly as it would if simply dropped, except it's also zipping side to side. Fall time = x ms. Distance D = x / 10. Bounces = (int) D. Location = D - (int) D; if Bounces is odd Location = 1 - Location;

 

[bilby]

There are two perfectly parallel vertical plates separated by 1 meter. Initial conditions a small ball at 10 meters high is starting away from one plate directly at the other plate.


Initial velocity is 100m/s. Ball mass is .01kg. Assume the system is perfect, the ball always travels in one plane when bouncing.

The plates and ball are hard with negligible loss of kinetic energy for the ball bouncing off the plates. Ignore air resistance.

1. how long before the ball hits the ground.
2. How many bounce between the plates.
3. Where between the plates does the ball hit the ground.

Ball mass is irrelevant. The ball falls exactly as it would if simply dropped, except it's also zipping side to side. Fall time = x ms. Distance D = x / 10. Bounces = (int) D. Location = D - (int) D; if Bounces is odd Location = 1 - Location;

The mass is irrelevant, but the radius of the ball is not, and as it is not given and cannot be derived, the problem cannot be solved. Your solution is correct for a ball of radius zero; Provision of the density of the ball would have allowed the radius to be calculated from the given mass. Given its other properties, it might be assumed to be made of unobtainium, however the density of unobtainium is currently unknown.

Of course, the actual experiment would replace the ball with a perfectly elastic, frictionless, spherical horse, of zero radius.

 

[skepticalbip]

the density of unobtainium is currently unknown.
.

Not so. The density of unobtainium is precisely what is required for its intended use... even if the application requires a negative density.

 

[steve_bank]

Obviously. To a first order approximation no different than a bullet, just distance folded by the plates. Mass was a red herring.

The old Myth Busters show actually dropped a bullet when a shot was fired and within measurement uncertainty both bullets hit the ground at the same time.



I did not work it out. Assuming perfect collisions of the ball and angle in = angle out I believe the ball should trace a folded parabolic arc to the ground.

 

[Swammerdami]

The bouncing ball puzzle reminded me of a way to derive the digits of pi from a ball-bouncing experiment. (Did I post on that back in early 2020?)

 

[Loren Pechtel]

There are two perfectly parallel vertical plates separated by 1 meter. Initial conditions a small ball at 10 meters high is starting away from one plate directly at the other plate.


Initial velocity is 100m/s. Ball mass is .01kg. Assume the system is perfect, the ball always travels in one plane when bouncing.

The plates and ball are hard with negligible loss of kinetic energy for the ball bouncing off the plates. Ignore air resistance.

1. how long before the ball hits the ground.
2. How many bounce between the plates.
3. Where between the plates does the ball hit the ground.

Ball mass is irrelevant. The ball falls exactly as it would if simply dropped, except it's also zipping side to side. Fall time = x ms. Distance D = x / 10. Bounces = (int) D. Location = D - (int) D; if Bounces is odd Location = 1 - Location;

The mass is irrelevant, but the radius of the ball is not, and as it is not given and cannot be derived, the problem cannot be solved. Your solution is correct for a ball of radius zero; Provision of the density of the ball would have allowed the radius to be calculated from the given mass. Given its other properties, it might be assumed to be made of unobtainium, however the density of unobtainium is currently unknown.

Of course, the actual experiment would replace the ball with a perfectly elastic, frictionless, spherical horse, of zero radius.

Oops! You're right, except it's a spherical cow!

 

[Shadowy Man]

The mass is irrelevant, but the radius of the ball is not, and as it is not given and cannot be derived, the problem cannot be solved. Your solution is correct for a ball of radius zero; Provision of the density of the ball would have allowed the radius to be calculated from the given mass. Given its other properties, it might be assumed to be made of unobtainium, however the density of unobtainium is currently unknown.

Of course, the actual experiment would replace the ball with a perfectly elastic, frictionless, spherical horse, of zero radius.

Oops! You're right, except it's a spherical cow!

If the radius is zero then the density is infinite!

If the mass is also zero then there is no gravitational force on it and it won’t fall.

So, I guess it has to be a black hole, not a ball.

 

[Jokodo]

A ball of infinite density would, however, pass right through the plates as if they were air.

 

[bilby]

A ball of infinite density would, however, pass right through the plates as if they were air.

The plates would need to be pretty tough even for a merely fairly high density projectile; 100ms^-1 is over 200mph, and is roughly the muzzle velocity of a black powder duelling pistol.