They like using special cases and it makes it easy to understand. Exactly the same applies to some and to slow. This is because they appear on *both sides* of the equation and you lose energy whatever combination you use. Try putting in some figures and it will be obvious.
"AL_n" wrote in news:Xns9E07C3EEA3C3Fzzzzzz@130.133.4.11:
PS - Sorry; I should have read the original post and watched the video. I was assuming this thread was about "anti-gravity" pseudo-science.
I thought I was being clever. :-7
Al
Um, I was there first.
Short answer : No, you don't need to provide the energy input you claim.
Longer answer : Thinking about the energy in the way you're doing will confuse things. I will admit I'm not happy about dealing with the fluid flows and energy going into them, and can't easily explain why your analysis is wrong, which is why I prefer to deal with forces and then add distance later. When you do that you discover your analysis must be wrong.
So, are you prepared to go with the forces?
Asked an answered. You claimed it takes 4 times as long to stop a car that's going 20 mph vs. one that's going 10 mph. My answer: no it doesn't. A normal car will take twice as long.
If Dennis will post his analysis, or give us a link to it, I'll be happy to show where he goes wrong.
Like you are prepared to go with the energy. if you can't understand something as simple as the energy transfers why should I think you get the forces correct? Anyway as I previously stated you can't do forces in isolation to energy, either both work or nothing does.
Here's a good forces one for you as someone introduced a wall earlier..
take a perfect wall, that is infinite, rigid and infinitesimally thin..
now take on air molecule from one side and put it on the other..
There you are infinite force.
I wonder what energy was used (well I don't but others will).
Rick Cavallaro
I assume the following, previously posted by dennis is what is being called his 'analysis'
********************************* Shame you don't read what I write, I have explained it before but it was ignored (not really a surprise, thin rick doesn't actually explain anything that is connected to the problem, he just invents something to deflect the issue and claims "its too hard for you to understand"..but its all about energy *and* thrust, if one doesn't work the other can be ignored.
Say you have a cart travelling at 20 m/s with a wind behind it blowing at 10 m/s
The relative wind at the cart is 10 m/s to the rear.
Now to stop the wind behind you have to accelerate the air going past the cart to 20 m/s.
Now put E=mv2 into the picture and work out how much energy it takes to accelerate the air going past from 10 m/s to 20 m/s so it stops the wind from behind.
My simple maths tells me its 4 times the energy you get from stopping the wind .
***************************
And there we have the proof you don't understand energy.
For anyone that cares..
the brakes use friction to turn kinetic energy to heat they have a maximum rate at which they can do this.. so as a car travelling twice as fast has four times the energy it takes four times as long to dissipate the heat in the brakes. So it takes four times as long to stop. It really is that simple.. rick doesn't have a clue, which is why he can't answer the energy questions.
Why not? Aren't they inextricably linked?
I'm only using momentum indirectly in order to get an expression for the propeller exit air speed V2 in terms of the parameters which JB's example gave (namely power in from wheels, thrust, car speed, wind speed, and hence V1 which is the relative wind speed seen by the car). The thrust figure implies (via the momentum equations) a particular inverse proportional relationship between "M dot" and "V2-V1", which I solve for V2 so that I can use the expression in the kinetic energy equations for the air before and after it has been through the prop.
I'm fully aware of that and don't have a problem with it. In the car frame, the air is being accelerated (backwards) from V1 to V2, and in the ground frame this corresponds to it being slowed by the same difference V2-V1. Perhaps it would be less misleading to say that in both frames the air is being decelerated with reference to the direction in which the car is moving, but while the deceleration in the ground frame is from a positive value to a less positive value (i.e. towards zero), in the car frame it is from a negative value to a more negative value (i.e. away from zero).
Thus the kinetic energy of the wind being decelerated actually increases in the car frame, while in the ground frame it decreases.
Yes, it could in principle be used to accelerate the car, but (and I may have misunderstood) I took JB's analysis to refer to a steady state car velocity, where all the acceleration of the car has already happened and the car is now at terminal velocity, with the energy loss from air resistance being what balances the energy harvested from the wind.
What conceptual problem? :-)
This brings to mind a conversation with my real ale pub landlord about the chevrons that mark some motorways that declare 'keep 2 chevrons apart'. To my mind, they are too close together, but he maintains that any modern car will stop within that distance as all the cars are slowing at the same rate.
I was bringing my wife back from a Scout training camp in the Lakes, when the small van in front of me sent smoke out from his tyres, Skidded out of the outside lane, into the middle lane, turned 90 degrees to the right and arrested his speed by driving slap bang into the crash barrier. Needles to say, I had to drive into him.
Dave
dennis@home
Hmmm -- I wonder what would happen if car manufacturers decided to size their brakes such that they could easily dissipate the heat from say a 100mph to 50mph panic stop. Those same brakes could then also easily dissipate the required amount of heat from 50mph to 0mph and thus the vehicle would only take twice the time to stop from the higher speed rather than four times. In this "new, enlightened manufacturer" scenario, the limit for typical panic stop would then not be heat dissipation at the brakes, but rather tire/pavement coefficient of friction.
Someone should build a car with brakes like this. Oh, wait -- EVERYONE DOES!!!!!!!!!!!!
You on the other hand must not have a car. Due to aerodynamic drag, on the braking limit your car will take less time to brake from 100mph to 50mph than it will from 50mph to 0mph. Try it with a stopwatch sometime.
They defy gravity they do not go down, they go up.
This one is a f***ing idiot
Balls!
I can't wait for the first manufacturer to get sued because he deliberately makes his brakes less efficient at low speeds just to satisfy your crazy idea about how brakes work.
In the real world the bit from 100 to fifty takes longer than the bit from
50-0 as anyone who has driven will be able to tell you.
Now we also have proof that thin air doesn't have a clue.
To think that anyone believes these people might have invented something that breaks the laws of physics amazes me, they can't even apply the simplest physics to the simplest cases like car brakes and get the correct answer. I blame the education system for not educating them.
My car isn't a brick and I specifically choose slow speeds to avoid aerodynamic effects you dumbass.
Doctor Drivel:
A lighter than air balloon is bowing to gravity rather than defying it. Gravity is actually forcing it UP.
JB
Well, you haven't actually pointed out where my forces bit is wrong yet, whereas I reckon your energy thing is wrong because it doesn't need to accelerate to 20ms-1.
Indeed. So, tell me where I've gone wrong with my forces.
True, but that's not reached in any modern brake. That's one of the reasons cars have ABS.
Therefore false.
False.
A modern car brakes at a fairly constant force, limited by tyre grip. v = u + at, a is the same, one of u and v is zero, therefore twice the speed = twice the time. Twice as long.
s = ut + 1/2 a*t*t t is doubled, s is quadrubled, therefore four times the distance.
That's o-level physics isn't it?
dennis@home
LOL -- the stopwatch doesn't lie. You should try one sometime.
That's a pretty neat trick how you avoid "aerodynamic effects". Care to share how slow you have to go to avoid those?
JB
He's right, they're not the full stopping distance apart, and they won't work if something suddenly stops faster than it normally can.
Balls. Cars have anti lock brakes because most drivers can't cope with difficult conditions and you are trying to deflect your inability to understand simple physics yet again.
For a car with cr@p tyre in the wet that might be true, but not any decent car, Oh I forget you don't do decent cars with proper suspension and tyres over there. Its funny that something like a Jag can pull 1.3G braking while you crappy cars only do about 0.8G, They both use similat tyres so I guess you don't push the pedal hard enough or you have cr@p brakes. Don't ever get close to a European car as you will go up his arse if he brakes.
Its primary school but the kids get it right unlike you.
"one of the reasons".
Not true. The limiting factor for a single stop for any modern car will be grip, not brakes. Show me evidence to the contrary if you have any. Once you're racing or descending mountains, life gets hotter, but that's not what we're talking about.
Where is "over there"? What sort of car do you think I drive? You might like to take a look at the address I use when posting to usenet and the groups I read for a clue to some of that answer.
Cite.
Why are my "crappy cars" any different to yours?
Cite.
I do try to avoid getting too close to people, but that's nothing to do with then being European. Of course if I'm driving I'll tend to be as close to a European car as you can get...
Nope.
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