- #71
Jurgen M
Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.
Relativistic mechanics is different, but you asked in the classical physics forum here.Jurgen M said:Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.
When an object with mass ##m## moves at speed ##v## it has non-physical "kinetic energy"Jurgen M said:I need this confirmation.
Generally I see increase in power at rocket as "mathematical manipulation" , I don't see anything useful in real physical sense in it.
It us essential that you understand the classical dynamics before you take on the appreciably hairier relativistic corrections. Otherwise you will never escape the confusion.Jurgen M said:Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.
Constant thrust doesn't equate to constant power.Jurgen M said:Let imagine that car with constant 500HP accelerate but resistance forces don't exist (aero drag,internal friction in engine and transmision,tyer rolling resistance etc etc..)
neglect fuel loss over time..
From 0-100km/h take in 4sec and burn 200mL petrol
Will car accelerate from 100-200km/h also in 4sec and burn 200mL petrol?
Here is how I look at it:
Car will also accelerate from 100-200km/h in 4sec and burn 200mL of petrol as well,because ΔV is same in both cases.
But
I know for constant acceleration car need constant Thrust.
I know that car engine produce constant power, Thurst = Power / Velocity, so thrust decrease as speed increase,that implies acceleration will be smaller from 100-200km/h,it will take longer then 4 sec and it will burn more then 200mL of petrol..or same thing from perspective of gears/torques:
Torque at wheel is what accelerate car,we can calculate thurst from wheel torque.
As car increase speed use higher gears which reduce torque at wheels,again thurst is reduced...So what is correct answer and what I am doing wrong?
There is thrust. The exhaust has zero energy, but when it was fuel traveling with the rocket, it had a very large amount of energy (since it was traveling 10km/s). Therefore, when the rocket engine exhausted the fuel, it removed 100% of the fuel's energy, and the rocket gained all the energy that was previously in the fuel. Since the rocket is lighter now, but still has just as much energy, it must be traveling faster.Jurgen M said:Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.
From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..
With constant thrust, the constant rate of Momentum Increase would be the relevant quantity. But even that will fail at high speeds over the ground.cmb said:If you car was powered by a constant thrust jet engine, you'd have been right.
More fuel per unit distance covered? No.Jurgen M said:now case with 2 constanat speeds, 100km/h and 200km/h
If aero drag don't exist, only exist tire resistance which is constant(lets say 50N) for all speeds, car will again use more fuel when driving at constant speed at 200km/h then on 100km/h?
Yes now make sense.Back to orignal acceleration question.jbriggs444 said:More fuel per unit distance covered? No.
More fuel per unit time taken? Yes.
That is going to depend on the rate of acceleration. We are assuming, per your stipulation, that there is negligible air resistance.Jurgen M said:Back to orignal acceleration question.
in acceleration from 100 to 200km/h car use more fuel per unit distance coverd than from 0 to 100km/h?
In my opinion, you have tried to leap in at the deep end of the topic. Cars, rockets and even electric motors are not ideal for starting on this problem.Jurgen M said:Kinetic energy raise with speed for rocket as well,but rocket(constant thrust) will have same acceleration and fuel burn from 0-100km/h and from 100-200km/h, so know that Ke raise with speed don't tell me nothing.
If we neglect fuel loss over time.
The most difference between car and rocket from what I see is that rocket has constant thrust and car has constant power(but reduced thrust with speed)..
reduced thrust implies reduced acceleration..
It seems all of you forget in this topic that car display consumtion in L/100km( fuel per unit of coverd distance) or maybe I need to note this.jbriggs444 said:That is going to depend on the rate of acceleration. We are assuming, per your stipulation, that there is negligible air resistance.
If the acceleration rate is held constant then the acceleration just amounts to an additional fixed force. The change in current speed will not have any effect on fuel consumption per unit distance.
If the acceleration rate is allowed to decline (perhaps engine power is limited) then the acceleration would amount to a large force at low speed and a smaller force at high speed. You would be more fuel efficient per unit distance at higher speed because of this.
We are assuming, perhaps counter-factually, that the engine has constant efficiency regardless of power setting. [My understanding is that real engines usually have a sweet spot where power is obtained most efficiently. Higher power output than that comes at the expense of increased fuel used per unit of energy released. Lower power output than that is also less efficient because the engine is wasting much of its output on the power required to just keep running]
You might concern yourself with the fact that kinetic energy goes with the square of velocity and it seems like you are burning 1 unit of energy to get to 100 km/h and 3 additional units of energy to get to 200 km/h. Yes indeed. But (for constant acceleration), you cover ##\frac{1}{2}at^2## of ground getting to 100 km/h and an additional ##\frac{3}{2}at^2## getting from 100 km/h to 200 km/h. That is a factor of three to one in distance which nicely matches the factor of three to one in kinetic energy
OK. So the assumption of constant acceleration is out the window. We have constant power instead.Jurgen M said:1.If someone ask me what is the main reason why every car accelerate significialy slower from 100-200 then from 0-100 if in both case deliver same power.
(we state here that car will behave the same even we neglect air,so aero drag is no reason,plus I knew that aero drag is not such a big to slow down acceleration so much,air is not water)
2. will car computer display higher L/100km if we accelarate from 100-200 then 0-100.
again with no aero drag
I can set my question in wacuum room.I.C. engine has gerabox which has gear ratios that keep engine full thortlle working allways between 6000rpm-8000rpm when lift weight.sophiecentaur said:Also, for these elementary models, you need slow speeds (a very few m/s) so you can at least omit air drag.
The diagram you show has a constant force device. However, you apparently want to use some gears or pulleys (not shown) so that the "50 N" thrust in first gear becomes, perhaps, a 10 N thrust in fifth gear.Jurgen M said:I can set my question in wacuum room.
1. Why weight acceleration is smaller when weight lift up from 100-200km/h then from 0-100km/h if drag force and weight is unchanged?
Yes, that explanation works.Jurgen M said:1.
For what I can see, tension in rope represent thrust for weight.
Logicaly tension in rope will be higher in 1gear then in 6gear,so engine reduce thrust as gears are changed,this I see as reason why weight reduced acceleration when lifitng up..
Let us not talk about rockets until we have wheeled propulsion squared away. Rockets conserve energy too. It is just that you have to look for the energy in different buckets.Jurgen M said:If rocket with constat thrust pull rope ,acceleration will be the same.
Yes.Jurgen M said:2.
L/100km will be smaller from 100-200km/h
Now I agree with everything, wish I before insist to use only fuel per units of distance,like cars allways show..jbriggs444 said:OK. So the assumption of constant acceleration is out the window. We have constant power instead.
I would answer 1 above by noting that for constant power, the work done per unit time is constant. The amount of work done per unit time is force times the distance covered in that unit time. That is to say that power ##P = \vec{F} \cdot \vec{v}##.
Increase velocity ##\vec{v}## and hold power ##P## constant and you'll need to reduce force ##\vec{F}##.
Acceleration ##\vec{a}## is given by ##\frac{\vec{F}}{m}##. Reduce propulsive force ##\vec{F}## and you've reduced forward acceleration ##\vec{a}##.
Or, if you prefer, you had to shift to a higher gear. Less available force at the wheels. The faster you go, the less acceleration results.
I would answer 2 by noting that your scenario is extraordinarily simple. The engine runs at constant power. Fuel consumption per unit time is fixed. You get the best fuel consumption per unit distance when you cover the greatest possible distance in that time. It is a direct proportionality. Fuel efficiency (per unit distance) is directly proportional to speed in this situation.
This is entirely independent of aerodynamic drag. Lots of drag, little drag. It is all the same. As long as you are running at a given constant power, all that matters for fuel efficiency is how fast you are going. The faster the better. Aerodynamic drag only enters in if it limits how fast you can go. Or if it forces you to use a bigger engine with a greater constant power output.
Not really. It is hard to do an apples to apples comparison when comparing wheeled propulsion to rocket propulsion.Jurgen M said:Then rocket enigne is better for drag racing, don't loose thrust with speed
You mean constant power ?jbriggs444 said:The diagram you show has a constant force device.
I.C. engine produce constant power only if you keep it at same rpm, but I assume that from 6000-8000rpm has aprox constant power...jbriggs444 said:This would not be a completely accurate representation of a constant power device, but adequately close.
You showed a 50 N weight on a string pulling a cart. That is a constant force device. It is not a constant power device.Jurgen M said:You mean constant power ?
If you are running your engine at max power (and if max power is not at red line), then the rate of change of power with respect to RPM must be zero there. So yes, if 7000 RPM is max power then 6000 to 8000 might be approximately max power.Jurgen M said:I.C. engine produce constant power only if you keep it at same rpm, but I assume that from 6000-8000rpm has aprox constant power...
No this not string pulling cart.jbriggs444 said:You showed a 50 N weight on a string pulling a cart. That is a constant force device. It is not a constant power device.
If you have constant 50 N force, it is a constant force device. Period. End Of Story.Jurgen M said:No this not string pulling cart.
Engine/gearbox stay at place and winding wheel lift up weight with rope like cran that has I.C. engine for lifting.Weight is accelerate up.
It has mass of 5kg, on the Earth this aprox 50N, and when accelerate up tension in rope is bigger then 50N...jbriggs444 said:If you have constant 50 N force, it is a constant force device. Period. End Of Story.
Maybe if it was a 5 kg mass, it could be constant power. But it's not 5 kg. It's labeled as 50 N.
Indeed bike can travel very fast if reduce aero drag, 183mph (295km/h) !PeroK said:I suspect that he doesn't ride a bicycle!
phinds said:A car can't have ANY acceleration if there is no resistance anywhere including the tire/road resistance because the wheels would just spin and the car would go nowhere.
Jurgen M said:Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.