OK, so here's one for you...

If I put my car into top gear and set the engine to 3,500 revolutions 
per minute, my car travels forward at almost exactly 75 miles per hour. 
So... how far forward does it travel for a single revolution of the engine??

Similarly, at that speed my car achieves roughly 50 miles per gallon of 
fuel. So how much fuel does it inject into each cylinder??

[My car has a four-cylinder, four-stroke internal compustion engine, and 
is normally-asperated. And runs on petrol, in case that wasn't obvious. 
Presumably the timing is therefore set so that exactly one cylinder 
fires for every revolution of the engine.]

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Invisible wrote:
> OK, so here's one for you...
> 
> If I put my car into top gear and set the engine to 3,500 revolutions
> per minute, my car travels forward at almost exactly 75 miles per hour.
> So... how far forward does it travel for a single revolution of the
> engine??
> 
> Similarly, at that speed my car achieves roughly 50 miles per gallon of
> fuel. So how much fuel does it inject into each cylinder??
> 
> [My car has a four-cylinder, four-stroke internal compustion engine, and
> is normally-asperated. And runs on petrol, in case that wasn't obvious.
> Presumably the timing is therefore set so that exactly one cylinder
> fires for every revolution of the engine.]
> 
Nope. Each cylinder fires once for every two revolutions. You have a 4
cyl engine therefore two cylinders fire for each rev.

John

-- 
I will be brief but not nearly so brief as Salvador Dali, who gave the
world's shortest speech. He said, "I will be so brief I am already
finished," then he sat down.

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Invisible wrote:
> OK, so here's one for you...
> 
> If I put my car into top gear and set the engine to 3,500 revolutions 
> per minute, my car travels forward at almost exactly 75 miles per hour. 
> So... how far forward does it travel for a single revolution of the 
> engine??

574643.

Ah, you want the unit? µm, naturally. This of course doesn't take count 
the "almost" you told, neither tires wearing nor possible sliding.

> Similarly, at that speed my car achieves roughly 50 miles per gallon of 
> fuel. So how much fuel does it inject into each cylinder??

None.

> [My car has a four-cylinder, four-stroke internal compustion engine, and 
> is normally-asperated. And runs on petrol, in case that wasn't obvious. 
> Presumably the timing is therefore set so that exactly one cylinder 
> fires for every revolution of the engine.]

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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On Fri, 16 May 2008 20:40:04 +0300, Eero Ahonen wrote:

> Ah, you want the unit? µm, naturally. This of course doesn't take count
> the "almost" you told, neither tires wearing nor possible sliding.

Or the effects of momentum. ;-)

Jim

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Invisible <voi### [at] devnull> wrote:
> OK, so here's one for you...
>
> If I put my car into top gear and set the engine to 3,500 revolutions
> per minute, my car travels forward at almost exactly 75 miles per hour.
> So... how far forward does it travel for a single revolution of the engine??

Is this a math puzzle, or a real question?

If it's a real question, it's almost impossible to answer from the information
given.  You have to take into account acceleration, momentum, friction, etc.

....Chambers

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>> If I put my car into top gear and set the engine to 3,500 revolutions
>> per minute, my car travels forward at almost exactly 75 miles per hour.
>> So... how far forward does it travel for a single revolution of the engine??
> 
> Is this a math puzzle, or a real question?
> 
> If it's a real question, it's almost impossible to answer from the information
> given.  You have to take into account acceleration, momentum, friction, etc.

How so?

The engine is directly coupled to the wheels. Unless the car looses 
traction, the forward motion of the car and the rotational speed of the 
wheels and engine all have a simple linear relationship.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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On Fri, 16 May 2008 21:11:34 +0100, Orchid XP v8 wrote:

>>> If I put my car into top gear and set the engine to 3,500 revolutions
>>> per minute, my car travels forward at almost exactly 75 miles per
>>> hour. So... how far forward does it travel for a single revolution of
>>> the engine??
>> 
>> Is this a math puzzle, or a real question?
>> 
>> If it's a real question, it's almost impossible to answer from the
>> information given.  You have to take into account acceleration,
>> momentum, friction, etc.
> 
> How so?
> 
> The engine is directly coupled to the wheels. Unless the car looses
> traction, the forward motion of the car and the rotational speed of the
> wheels and engine all have a simple linear relationship.

There isn't a linear relationship between the rotational speed of the 
wheels and the speed of the engine.

If I take my foot off the gas, the engine revs drop.  The car may or may 
not slow down (depending on the road and the landscape).

Similarly, if I apply the brakes to a full stop, the engine doesn't die, 
it idles.

Then of course if you put the car in neutral, you've completely 
disengaged the engine from the wheels.

Jim

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Invisible <voi### [at] devnull> wrote:
> If I put my car into top gear and set the engine to 3,500 revolutions 
> per minute, my car travels forward at almost exactly 75 miles per hour. 
> So... how far forward does it travel for a single revolution of the engine??

  Isn't this laughably simple to calculate?

  3500 rpm = 3500*60 rph, and in 1 hour it travels 75 miles. Thus divide
75 miles by 3500*60 revolutions, and you have the amount of miles per one
revolution, which would be approximately 0.0003571428 miles.

  Given that I'm not sure which whacky units miles are divided into,
I'll just convert those to more understandable units, so 75/(3500*60)
miles is approximately 57.5 centimeters.

-- 
                                                          - Warp

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Jim Henderson wrote:

> There isn't a linear relationship between the rotational speed of the 
> wheels and the speed of the engine.

Well that's news.

> If I take my foot off the gas, the engine revs drop.  The car may or may 
> not slow down (depending on the road and the landscape).

Not true. The engine revs depend on the speed of the vehicle.

If you don't believe me, try rolling down a hill at 30, and then stick 
the car into 1st gear. Watch the revs go through the roof, even though 
you haven't touched the gas.

> Similarly, if I apply the brakes to a full stop, the engine doesn't die, 
> it idles.

Um, yes it does. Unless...

> Then of course if you put the car in neutral, you've completely 
> disengaged the engine from the wheels.

...you disenguage the engine from the wheels. Then there really isn't 
any relationship. But as long as the car stays in the same gear, there 
is a fixed relationship.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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>> If I put my car into top gear and set the engine to 3,500 revolutions 
>> per minute, my car travels forward at almost exactly 75 miles per hour. 
>> So... how far forward does it travel for a single revolution of the engine??
> 
>   Isn't this laughably simple to calculate?

Yes - *if* you can do things like compute 60/75 *mentally*. :-P And also 
if you know how many yards there are in a mile, and all kinds of other 
random stuff.

[I found myself wondering about this in the car on the way home.]

Now, if you have access to a *computer* - or even better, Google - then 
yes, it becomes quite easy.

>   Given that I'm not sure which whacky units miles are divided into,
> I'll just convert those to more understandable units, so 75/(3500*60)
> miles is approximately 57.5 centimeters.

That's the number I arrived at too.

Google also tells me that 75 mph = 33 m/s. Which frankly seems really 
damned fast. (!) I mean, my car is only, like, 4 m long or something... 
That means the car covers almost 10x its own length in 1 second! o_O

No wonder hitting anything is an instant fatality...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Jim Henderson wrote:
> There isn't a linear relationship between the rotational speed of the 
> wheels and the speed of the engine.

There is, in a (relatively primitive) manual transmission. There is not, 
however, a linear relationship between engine RPM and gasoline usage.

> Similarly, if I apply the brakes to a full stop, the engine doesn't die, 
> it idles.

When did you last drive a manual transmission? :-)

-- 
   Darren New / San Diego, CA, USA (PST)
     "That's pretty. Where's that?"
          "It's the Age of Channelwood."
     "We should go there on vacation some time."

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On Fri, 16 May 2008 15:20:56 -0700, Darren New wrote:

> Jim Henderson wrote:
>> There isn't a linear relationship between the rotational speed of the
>> wheels and the speed of the engine.
> 
> There is, in a (relatively primitive) manual transmission. There is not,
> however, a linear relationship between engine RPM and gasoline usage.

How many cars use a transmission that would count as "relatively 
primitive"?  But I do agree that there's not a linear relationship 
between RPM and MPG.

>> Similarly, if I apply the brakes to a full stop, the engine doesn't
>> die, it idles.
> 
> When did you last drive a manual transmission? :-)

D'oh!  I was thinking automatic - I learned to drive on a stick....

Jim

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On Fri, 16 May 2008 21:27:22 +0100, Orchid XP v8 wrote:

> Jim Henderson wrote:
> 
>> There isn't a linear relationship between the rotational speed of the
>> wheels and the speed of the engine.
> 
> Well that's news.
> 
>> If I take my foot off the gas, the engine revs drop.  The car may or
>> may not slow down (depending on the road and the landscape).
> 
> Not true. The engine revs depend on the speed of the vehicle.
> 
> If you don't believe me, try rolling down a hill at 30, and then stick
> the car into 1st gear. Watch the revs go through the roof, even though
> you haven't touched the gas.
> 
>> Similarly, if I apply the brakes to a full stop, the engine doesn't
>> die, it idles.
> 
> Um, yes it does. Unless...
> 
>> Then of course if you put the car in neutral, you've completely
>> disengaged the engine from the wheels.
> 
> ...you disenguage the engine from the wheels. Then there really isn't
> any relationship. But as long as the car stays in the same gear, there
> is a fixed relationship.

Perhaps in a manual transmission - in my automatic, though, I'll stick 
with what I indicated....I s'pose I could ask my father-in-law when I 
talk to him next, since his life's work has been working on cars....

Jim

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Jim Henderson wrote:
> Perhaps in a manual transmission - in my automatic, though, I'll stick 
> with what I indicated.

Sure. But the automatic has a clutch built into it. That's why you only 
have two pedals.

-- 
   Darren New / San Diego, CA, USA (PST)
     "That's pretty. Where's that?"
          "It's the Age of Channelwood."
     "We should go there on vacation some time."

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Jim Henderson wrote:
> How many cars use a transmission that would count as "relatively 
> primitive"? 

One that has distinct manually-shifted gears without synchromesh. Lots 
of newer cars have things like variable transmissions.

-- 
   Darren New / San Diego, CA, USA (PST)
     "That's pretty. Where's that?"
          "It's the Age of Channelwood."
     "We should go there on vacation some time."

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On Fri, 16 May 2008 16:06:35 -0700, Darren New wrote:

> Jim Henderson wrote:
>> Perhaps in a manual transmission - in my automatic, though, I'll stick
>> with what I indicated.
> 
> Sure. But the automatic has a clutch built into it. That's why you only
> have two pedals.

True....

Jim

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On Fri, 16 May 2008 16:08:15 -0700, Darren New wrote:

> Jim Henderson wrote:
>> How many cars use a transmission that would count as "relatively
>> primitive"?
> 
> One that has distinct manually-shifted gears without synchromesh. Lots
> of newer cars have things like variable transmissions.

Yeah, that's what I'm thinking of, just didn't know the name of it...

Jim

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>>   Given that I'm not sure which whacky units miles are divided into,
>> I'll just convert those to more understandable units, so 75/(3500*60)
>> miles is approximately 57.5 centimeters.
>
> That's the number I arrived at too.

Sure, 1.885 feet per revolution.

> Google also tells me that 75 mph = 33 m/s. Which frankly seems really 
> damned fast. (!) I mean, my car is only, like, 4 m long or something... 
> That means the car covers almost 10x its own length in 1 second! o_O
>
> No wonder hitting anything is an instant fatality...

Hmm, yeah, 110 feet per second, wear your seatbelt.

0.0027 tsp of gas per cylinder. (other factors ignored).

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Darren New wrote:
> Jim Henderson wrote:
>> Perhaps in a manual transmission - in my automatic, though, I'll stick 
>> with what I indicated.
> 
> Sure. But the automatic has a clutch built into it. That's why you only 
> have two pedals.
> 

http://en.wikipedia.org/wiki/Variomatic

just plugging dutch technology that never really caught on.

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> Perhaps in a manual transmission - in my automatic, though, I'll stick 
> with what I indicated...

Oh, well, yeah. Automatic transmission basically means it can change 
gear and lock/unlock the clutch at will. And doing both of those things 
changes the relationship. ;-)

My car has manual transmission.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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>> There isn't a linear relationship between the rotational speed of the 
>> wheels and the speed of the engine.
> 
> There is, in a (relatively primitive) manual transmission.

Indeed.

> There is not, 
> however, a linear relationship between engine RPM and gasoline usage.

No - that would depend on the smoothness of the road, the inclination, 
wind speed, and a bunch of other factors. Even so, I would expect it to 
be possible to come up with an *approximate* number for fuel used per 
cylinder.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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andrel wrote:
> 
> http://en.wikipedia.org/wiki/Variomatic
> 
> just plugging dutch technology that never really caught on.

It did, when improved enough.

http://www.audiworld.com/news/99/multitronic/content.shtml

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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>> No wonder hitting anything is an instant fatality...
> 
> Hmm, yeah, 110 feet per second, wear your seatbelt.

And won't walk across busy motorways at night while dressed only in 
black. ;-)

Also, DON'T HIT A TREE!

[A nice mature Oak has brunches going up to 30 feet into the air, and 
roots going down to 30 feet into the ground. The entire organism 
probably weighs multiple tonnes. If you hit it, IT'S NOT GOING ANYWHERE!]

> 0.0027 tsp of gas per cylinder. (other factors ignored). 

Hmm, that's actually really small... that's like... 0.015 ml. o_O

That can't be right, surely? That's less than one droplet per cylinder! 
How would you portion out such a tiny amount? Surface tension would seem 
to make this impossible.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Orchid XP v8 wrote:
> 
> No - that would depend on the smoothness of the road, the inclination, 
> wind speed, and a bunch of other factors. Even so, I would expect it to 
> be possible to come up with an *approximate* number for fuel used per 
> cylinder.
> 

Yes, ~1/4 of used fuel is used for one cylinder (but not injected in one 
- it's injected to the air intake before cylinders;). Just multiply the 
~28,75cm with consumption (l/100km) and you'll get the average amount of 
fuel .

Just a side-note: given that IIRC your car is <2l naturally aspirated, 
I'd guess the maximum possible fuel used per minute per cylinder with 
original parts is somewhere around 200cc (the injector capacity), giving 
total of 0,8l/min. OTOH, that much is used rarely, basically only when 
revving with pedal on the floor at the rpm of maximum power.

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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Orchid XP v8 wrote:
> 
>> 0.0027 tsp of gas per cylinder. (other factors ignored). 
> 
> Hmm, that's actually really small... that's like... 0.015 ml. o_O

On one rev the car moves 28,75cm, in that time 2 sparks are launched. 
Google says 40mpg (US, AFAIK it's pretty near 50imperial MPG) converts 
to 5,8l/100km (which is pretty low for 120km/h, but possible, if you 
have a modern diesel with long gearbox and good aerodynamics). 
5,8l/100km makes 
0,058l/km=0,000058l/m=0,00003335l/rev=0,00001668l/spark=0,01668ml/spark. 
Yep, we're on the same line - I'd say the differences comes from the 
roundings and converts.

> That can't be right, surely? That's less than one droplet per cylinder! 
> How would you portion out such a tiny amount? Surface tension would seem 
> to make this impossible.

To get to that consumption on those speeds you'll need a modern diesel 
engine (or a hybrid-system), for example common rail -engine. In those, 
the diesel is packed with very high pressure (>1000bar) on a kind of 
pipe before the cylinders and released to cylinders very precisely with 
piezo-electric injectors. Really fine mechanics, think about a 
manufacturing fault in there...

http://en.wikipedia.org/wiki/Common_rail

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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Eero Ahonen wrote:
> On one rev the car moves 28,75cm, in that time 2 sparks are launched. 

HUGE mistake - on HALF rev the car moves 28,75cm and in that time only 1 
spark is launched, therefore...

> Google says 40mpg (US, AFAIK it's pretty near 50imperial MPG) converts 
> to 5,8l/100km (which is pretty low for 120km/h, but possible, if you 
> have a modern diesel with long gearbox and good aerodynamics). 
> 5,8l/100km makes 
> 0,058l/km=0,000058l/m=0,00003335l/rev=0,00001668l/spark=0,01668ml/spark. 
> Yep, we're on the same line - I'd say the differences comes from the 
> roundings and converts.

... these last numbers DO NOT MATCH, the correct final number being 
0,03335ml/spark.

> To get to that consumption on those speeds you'll need a modern diesel 
> engine (or a hybrid-system), for example common rail -engine. In those, 

Also to be noted, if you really have common-rail diesel or something 
similar (I recalled you have a non-GDI petrol engine), the fuel actually 
is injected to the cylinder, not to the intake.

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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Invisible wrote:
> OK, so here's one for you...
> 
> If I put my car into top gear and set the engine to 3,500 revolutions 
> per minute, my car travels forward at almost exactly 75 miles per hour. 
> So... how far forward does it travel for a single revolution of the 
> engine??
> 
> Similarly, at that speed my car achieves roughly 50 miles per gallon of 
> fuel. So how much fuel does it inject into each cylinder??
> 
> [My car has a four-cylinder, four-stroke internal compustion engine, and 
> is normally-asperated. And runs on petrol, in case that wasn't obvious. 
> Presumably the timing is therefore set so that exactly one cylinder 
> fires for every revolution of the engine.]
> 

The fuel consumption of a vehicle is often quoted (in metric) as litres 
per 100km.  Suppose a medium sized sedan gives 10L/100km to keep the 
maths easy.

So we have fuel consumption as a volume per distance.  That means it can 
be reduced to an area.  Think of it as the cross-section area of a 
cylindrical line of fuel that the vehicle is consuming as it hurtles 
down the road.  Of course when you accelerate the engine guzzles more 
than this average and while cruising at some optimal speed (around 
80km/hour ?) it would sip a bit less so the shape is tapered.  It isn't 
a simple cylinder but rather a lumpy one.

Lets ignore these variations and work out how large this cross-section 
is on average.

Well do the calculation and you get 0.1 mm^2.  That is one tenth of a 
square millimeter!  Depending on the size and resolution of your screen 
and the font size that you use it is about the area of a full stop.

So as you cruise along a motorway your vehicle is slurping a stream of 
fuel about the same size across as a largish grain of sand, burning it 
to produce hot gases and farting them out the rear.

I find the fact that such a seemingly miniscule amount of fuel can be 
used to propel quite massive vehicles along at such speeds quite 
amazing.  The energy density and convenience of petroleum fuels is just 
astonishing.  As someone once said if you were to write down the ideal 
properties of a vehicle fuel you would pretty much define petrol.  High 
energy density but not (terribly) explosive, liquid over a wide range of 
temperatures, cheap (well relatively), easy to handle, not (too) toxic 
or corrosive etc.

Second thought though is that when you add up the consumption of tens of 
thousands of vehicles along a section of major road each day it does 
start to amount to quite a lot.  Those fractions of a square millimeter 
accumulate to something like the cross section of a fire hose or water 
main.  Imagine the threads of each vehicle weaving with others to form a 
massive cable down the center of the road.  Smaller roads might equate 
to a garden hose each.  Sum this across a city and it is a huge web 
spread down every road and tangled across every intersection.

Over the course of a year the sum cross section along a major road would 
be larger than the cross section of a petrol tanker truck.  So it is 
like every highway and major road being lined with tankers parked nose 
to tail.

And all of this is with a liquid that was produced millions of years 
ago.  It is the captured sunlight that fell on forests in the time of 
the dinosaurs.

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> I find the fact that such a seemingly miniscule amount of fuel can be 
> used to propel quite massive vehicles along at such speeds quite 
> amazing.  The energy density and convenience of petroleum fuels is just 
> astonishing.

On the other hand, steam engines are more efficient. (I.e., transform 
more of the available energy into useful work.) But they're really not 
"convinient" at all.

> Second thought though is that when you add up the consumption of tens of 
> thousands of vehicles along a section of major road each day it does 
> start to amount to quite a lot.

Uh, yeah...

> And all of this is with a liquid that was produced millions of years 
> ago.  It is the captured sunlight that fell on forests in the time of 
> the dinosaurs.

Indeed.

Still, you'd think it would be possible to somehow take plants living 
today and transform their tissues into some kind of highly flammable 
hydrocardon with reasonably efficiency. This would seem the obvious 
sustainable solution.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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"Orchid XP v8" <voi### [at] devnull> a écrit dans le message de news: 
482fe97c$1@news.povray.org...
> On the other hand, steam engines are more efficient. (I.e., transform more 
> of the available energy into useful work.) But they're really not 
> "convinient" at all.
Stirling engine generator >>> electricity powered wheels...
http://en.wikipedia.org/wiki/Stirling_engine
>
>> Second thought though is that when you add up the consumption of tens of 
>> thousands of vehicles along a section of major road each day it does 
>> start to amount to quite a lot.
>
> Uh, yeah...
>
>> And all of this is with a liquid that was produced millions of years ago. 
>> It is the captured sunlight that fell on forests in the time of the 
>> dinosaurs.
>
> Indeed.
>
> Still, you'd think it would be possible to somehow take plants living 
> today and transform their tissues into some kind of highly flammable 
> hydrocardon with reasonably efficiency. This would seem the obvious 
> sustainable solution.
There are researches on hydrocarbon synthesis directly from CO2 and H2.
This could be a better solution because arid (but close to the sea )and 
sunny (solar energy) zones could be used instead of fertile lands better for 
food production.

Marc

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Warp wrote:
...
>   3500 rpm = 3500*60 rph, and in 1 hour it travels 75 miles. Thus divide
> 75 miles by 3500*60 revolutions, and you have the amount of miles per one
> revolution, which would be approximately 0.0003571428 miles.
> 
>   Given that I'm not sure which whacky units miles are divided into,
...

0.0003571428 mile
= 1 foot + 2 hands + 2.6285678 inches
= 22 and 5/8 inch (and something).

http://www.google.com/search?q=1+foot+2+hands+2.6285678+inches+in+miles
http://www.google.com/search?q=%2822+5/8+%29+inches+in+miles

1 mile
= 8 furlongs
= 80 chains
= 1760 yards
= 8000 links
= 5280 feet
= 15840 hands
= 63360 inches

1 furlong
= 10 chains
= 220 yards
= 660 feet
= 1000 links
= 7920 inches

1 chain
= 22 yards
= 66 feet
= 100 links
= 198 hands
= 792 inches

1 yard
= 3 feet
= 9 hands
= 36 inches

1 foot
= 3 hands
= 12 inches

1 hand
= 4 inches

1 inch
= 1000 mils

-- 
Tor Olav
http://subcube.com

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Tor Olav Kristensen wrote:

> 1 mile
> = 8 furlongs
> = 80 chains
> = 1760 yards
> = 8000 links
> = 5280 feet
> = 15840 hands
> = 63360 inches
> 
> 1 furlong
> = 10 chains
> = 220 yards
> = 660 feet
> = 1000 links
> = 7920 inches
> 
> 1 chain
> = 22 yards
> = 66 feet
> = 100 links
> = 198 hands
> = 792 inches
> 
> 1 yard
> = 3 feet
> = 9 hands
> = 36 inches
> 
> 1 foot
> = 3 hands
> = 12 inches
> 
> 1 hand
> = 4 inches
> 
> 1 inch
> = 1000 mils

THANK GOD FOR METRIC!!! >_<

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Tor Olav Kristensen wrote:
> 
> 1 mile
> = 8 furlongs
> = 80 chains
> = 1760 yards
> = 8000 links
> = 5280 feet
> = 15840 hands
> = 63360 inches
> 
-clip-
> 
> 1 inch
> = 1000 mils
> 

I think I'll stay preferring the metric system.

-- 
Eero "Aero" Ahonen
    http://www.zbxt.net
       aer### [at] removethiszbxtnetinvalid

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Orchid XP v8 wrote:
> THANK GOD FOR METRIC!!! >_<
> 

What he said.

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On Sun, 18 May 2008 18:44:19 +0200, Tor Olav Kristensen
<tor### [at] TOBEREMOVEDgmailcom> wrote:

>
>1 mile
>= 8 furlongs
>= 80 chains
>= 1760 yards

Where were you when I was at school :-)

Your post looks like the back of my old school jotters :-)

Yes we had to learn all that stuff by heart.
-- 

Regards
     Stephen

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> The engine is directly coupled to the wheels.

Assuming the clutch is engaged, yes this is pretty much true, the only tiny 
difference can come from twisting in the driveshafts and bending of the 
teeth on the gears.  So it's pretty much coupled :-)

> Unless the car looses traction, the forward motion of the car and the 
> rotational speed of the wheels and engine all have a simple linear 
> relationship.

Except pretty much the whole time, part of the tyre has lost traction with 
the road, if it hadn't then it would mean there were no forces being 
generated, and you wouldn't go anywhere.

There's a quantity called "slip ratio", which is the ratio between the 
actual speed the wheel is rotating, and what it "should" be (ie vehicle 
speed divided tyre rolling circumference).  The slip ratio (SR) is never 
exactly 1, only when you are applying no forces to the wheel (ie no power 
and no braking).  If you brake moderately hard, it can easily go down to 
0.8, and under normal acceleration can be 1.05 or so depending on your 
tyres.

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>> There is, in a (relatively primitive) manual transmission. There is not,
>> however, a linear relationship between engine RPM and gasoline usage.
>
> How many cars use a transmission that would count as "relatively
> primitive"?  But I do agree that there's not a linear relationship
> between RPM and MPG.

I'd say that about 90% or more of cars sold here in Europe the transmission 
is a "relative primitive" manual.  The reason is for efficiency, any extra 
junk in the transmission (like fancy fluid couplings or multiple-clutches 
etc) just increases fuel consumption.  You connect the engine as directly to 
the wheels as you possibly can, otherwise you're just wasting energy.

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> If I put my car into top gear and set the engine to 3,500 revolutions per 
> minute, my car travels forward at almost exactly 75 miles per hour. So... 
> how far forward does it travel for a single revolution of the engine??

I'll assume the other answers are correct, 0.57 meters.

> Similarly, at that speed my car achieves roughly 50 miles per gallon of 
> fuel. So how much fuel does it inject into each cylinder??

50 miles per gallon = 5.6e-5 litres per meter according to google.

Which is about 3.22e-5 litres per revolution.

Let's use units of mass here, because I think it's easier to imagine the 
amount, so we have about 24 mg of petrol being used per revolution.

So in your engine, that means 12 mg of fuel is being injected each time into 
each cylinder (the fuel is injected every 2nd revolution into each 
cylinder).

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> I find the fact that such a seemingly miniscule amount of fuel can be used 
> to propel quite massive vehicles along at such speeds quite amazing.

And they are improving all the time - I'm pretty sure that in my new car 
I'll be able to do 1000 km on a single 50 litre tank if I take it easy (just 
got 780 km on my first tank, but a lot of that was going over 160 km/hr!). 
Even 5 years ago that would have been impossible in most cars.

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Paul Fuller <pgf### [at] optusnetcomau> wrote:
> I find the fact that such a seemingly miniscule amount of fuel can be 
> used to propel quite massive vehicles along at such speeds quite 
> amazing.

  Once the car is moving it only requires the amount of fuel necessary
to overcome the friction which would otherwise stop the car. Given that
the car has an enormous amount of momentum, the amount of energy needed
to keep it against such small things as air friction isn't all that large.

  Compare it to the amount of fuel needed to get the car from 0 to 100 km/h
in a relatively short amount of time.

-- 
                                                          - Warp

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> to keep it against such small things as air friction isn't all that large.

At 75 mph air friction isn't small.  A lot of small cars give up not much 
faster than this because their engine cannot generate enough power to 
overcome the air friction.

>  Compare it to the amount of fuel needed to get the car from 0 to 100 km/h
> in a relatively short amount of time.

Much less than getting to 100 km/h in a long time...

IMO, the surprise at the small amount of fuel needed comes from the fact 
that the energy density of petrol is so high.  We hear figues like 6 litres 
per 100 km, but 6 litres of petrol equals around 200 MJ of energy, enough to 
lift a 747 plane 100 metres in the air!

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Orchid XP v8 wrote:
> 
> Still, you'd think it would be possible to somehow take plants living 
> today and transform their tissues into some kind of highly flammable 
> hydrocardon with reasonably efficiency. This would seem the obvious 
> sustainable solution.
> 

They do something of the sort - Biodiesel
But that's what politicians are blaming on the food shortages and price 
increases.  Go figure....



Tom

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> They do something of the sort - Biodiesel
> But that's what politicians are blaming on the food shortages and price 
> increases.  Go figure....

And for screwing up engines.

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How about this classical physics problem:

  You drop a rock into a water well. Then you hear the splash. Exactly
10 seconds pass between you letting go of the rock and when you hear the
splash. How deep is the well?

  We can assume that air friction is negligible, that g = 9.81 m/s^2,
and that the speed of sound is 340 m/s.

  Don't just give the answer, also write your whole reasoning.

-- 
                                                          - Warp

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Warp wrote:
>   How about this classical physics problem:
> 
>   You drop a rock into a water well. Then you hear the splash. Exactly
> 10 seconds pass between you letting go of the rock and when you hear the
> splash. How deep is the well?
> 
>   We can assume that air friction is negligible, that g = 9.81 m/s^2,
> and that the speed of sound is 340 m/s.
> 
>   Don't just give the answer, also write your whole reasoning.

Hmm, takes me back to school.

10 = t(sound) + t(rockfall)

t(sound) = x/340 where x is the well depth

x = 1/2 * g * t(rockfall)^2
so
t(rockfall) = sqrt(2x/g)

substituting back in,

10 = x/340 + sqrt(2x/g)

sub y = sqrt(x), rearrange

(1/340)*y^2 + sqrt(2/g)*y - 10 = 0

solve quadratic for positive y, square y to get x

I get x = 393.25m, sounds about right but my arithmetic might be wonky.

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Warp wrote:
>   How about this classical physics problem:
> 
>   You drop a rock into a water well. Then you hear the splash. Exactly
> 10 seconds pass between you letting go of the rock and when you hear the
> splash. How deep is the well?
> 
>   We can assume that air friction is negligible, that g = 9.81 m/s^2,
> and that the speed of sound is 340 m/s.
> 
>   Don't just give the answer, also write your whole reasoning.

Hmm, the object has an initial velocity v0 = 0 m/s.

Assuming zero air resistence, the object as infinite terminal velocity 
and accelerates uniformly at 9.81 m/s^2. This yields

   vt = 9.81 t

The integral of this would be the object's position:

   pt = 9.81/2 t^2 = 4.905 t^2

We hear the splash at t=10. Assuming that sound is initiated at time t=K 
and it propogates towards the observer at 340 m/s, we have

   K = 10 - d/340

where d is the distance to the bottom of the well.

So, we know that d must obey

   K = 10 - d/340   and   4.905 K^2 = d

Rearranging, we have

   K - 10 = -d/340
   340 (K - 10) = -d
   340 K - 3,400 = -d
   -340 K + 3,400 = d

So d = -340 K + 3,400 = 4.905 K^2.

At this point, I have insufficient knowledge of algebra to deduce a 
solution, although obviously one exists.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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> So d = -340 K + 3,400 = 4.905 K^2.
>
> At this point, I have insufficient knowledge of algebra to deduce a 
> solution, although obviously one exists.

You should learn how to calculate the solution to equations like this, they 
come up everywhere:

http://en.wikipedia.org/wiki/Quadratic_equation

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>> So d = -340 K + 3,400 = 4.905 K^2.
>>
>> At this point, I have insufficient knowledge of algebra to deduce a 
>> solution, although obviously one exists.
> 
> You should learn how to calculate the solution to equations like this, 
> they come up everywhere:
> 
> http://en.wikipedia.org/wiki/Quadratic_equation

Solving a single quadratic equation is trivial. Solving a pair of linear 
equations is trivial. But I have no idea how to solve a pair of 
equations where one is linear and the other isn't.

[Actually, I can still remember sitting down and deriving the quadratic 
formula from first principles. It's quite instructive! Whoever it was 
that figured out that A*B - 4*(A+B) = A-B is some kind of genius if you 
ask me... The next day I tried to do the same with a cubic, but... uh... 
failed.]

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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>>> So d = -340 K + 3,400 = 4.905 K^2.
>>>
>>> At this point, I have insufficient knowledge of algebra to deduce a 
>>> solution, although obviously one exists.

...

> Solving a single quadratic equation is trivial.

What do you call that equation you got at the top there?

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scott wrote:
>>>> So d = -340 K + 3,400 = 4.905 K^2.
>>>>
>>>> At this point, I have insufficient knowledge of algebra to deduce a 
>>>> solution, although obviously one exists.
> 
> ...
> 
>> Solving a single quadratic equation is trivial.
> 
> What do you call that equation you got at the top there?

A pair of equations welded together?

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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>>>>> So d = -340 K + 3,400 = 4.905 K^2.

> A pair of equations welded together?

A = B = C implies that B = C, no?

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