What speed v(t) enables constant power by release compressed spring?

[cairoliu]

We know linear spring force F = kx(t), k = spring constant, during any moment t of energy release.
displacement x(t) = ʃv(t)dt
the power p = F*V = kx(t)v(t)= kv(t)* ʃv(t)dt
My question:
Is there a mathematical function of special v(t) to make power p constant?

 

[cairoliu]

at moment t = 0, fully compressed spring x(0) = L, the max, and F is also max
at moment t = T, fully released spring x(T) = 0, F = 0
physically, I wander how to retard the spring to get constant power.
Minor amendment to the earlier message:
displacement x(t) = L - ʃv(t)dt
the power p = F*V = kx(t)v(t)= kv(t)* (L- ʃv(t)dt)
if v(t) is constant c, then p = kc(L-c*t), it means power will linearly decrease.

 

[cairoliu]

nobody interested?
Force/load retardation against energy source is an art to get constant power!

 

[jack action]

I was watching this thread, hoping to see what others had to say.

All I could determine was that $\dot{x}(L-x) = \text{constant}$ or:
$$\frac{\partial}{\partial t}\left(\dot{x}(L-x)\right) = 0$$
$$(L-x)\frac{\partial}{\partial t}\left(\dot{x}\right) + \dot{x}\frac{\partial}{\partial t}\left(L-x\right) = 0$$
$$(L-x)\ddot{x} - \dot{x}^2 = 0$$
Therefore, you function $v(t)$ would be:
$$v(t) = \dot{x} = \sqrt{(L-x)\ddot{x}}$$
According to Wolfram the following functions $x(t)$ would satisfy this condition:

 

[Lnewqban]

You could use a mechanism to make the spring (or any other elastic source of force) output more linear regarding force and work.
However, it is not achievable for power, I believe, because the output speed (linear or rotational) will always suffer, even when the force or torque could be kept more or less constant.

You could however, accumulate the potential energy of the work done by the spring, in order to used later at constant power.
Let's say, your spring pumps water into a reservoir for later use in a turbine.

 

[cairoliu]

I was watching this thread, hoping to see what others had to say.

All I could determine was that $\dot{x}(L-x) = \text{constant}$ or:
$$\frac{\partial}{\partial t}\left(\dot{x}(L-x)\right) = 0$$
$$(L-x)\frac{\partial}{\partial t}\left(\dot{x}\right) + \dot{x}\frac{\partial}{\partial t}\left(L-x\right) = 0$$
$$(L-x)\ddot{x} - \dot{x}^2 = 0$$
Therefore, you function $v(t)$ would be:
$$v(t) = \dot{x} = \sqrt{(L-x)\ddot{x}}$$
According to Wolfram the following functions $x(t)$ would satisfy this condition:

View attachment 281457
View attachment 281458​

so by applying boundary conditions, I get x(t)= L(1-(t/T)^0.5)
v(t) = dx(t)/dt = -0.5L*T^0.5*t^(-0.5)
But it seems that power P(t) still a function of time.

 

[cairoliu]

Also wish get constant power from a compressed air cylinder, is it analogy to release energy in compressed spring?
Having the answer to the said spring problem, may help fix how to isothermally control the piston displacement in cylinders of Compressed Air Energy Storage system.

 

[cairoliu]

so by applying boundary conditions, I get x(t)= L(1-(t/T)^0.5)
v(t) = dx(t)/dt = -0.5L*T^0.5*t^(-0.5)
But it seems that power P(t) still a function of time.

I made mistake.
It should be:
v(t) = dx(t)/dt = -0.5L*(T*t)^(-0.5)
so P(t) = k*(L- x)*v(t) = -(0.5*k*L^2)/T, constant now.
And 0.5*k*L^2 is just the total energy stored in the spring.
so averaged by the retardation time T, it does get constant power. Really make physical sense!
The negative speed means spring bounce load back -x direction, and in fact I mistakingly use the +x dirction for the bounce force; if I fix it, then power is correctively positive.
The 2nd solution

cannot meet boundary conditions, so it is abandoned.
Good job, dear Jack Action!

 

[cairoliu]

You could use a mechanism to make the spring (or any other elastic source of force) output more linear regarding force and work.
However, it is not achievable for power, I believe, because the output speed (linear or rotational) will always suffer, even when the force or torque could be kept more or less constant.

You could however, accumulate the potential energy of the work done by the spring, in order to used later at constant power.
Let's say, your spring pumps water into a reservoir for later use in a turbine.

View attachment 281459

View attachment 281460

Sounds physically OK, but dodges math problem.
In engineering, I think, it needs flywheel and feedback to make it work, unlike Jack's solution without temporary energy buffering and recovering by feedback loop.
Engineering Jack's solution is not easy too, it may need computer assistance to accurately control retardation for the calculated real time displacement.

 

[sophiecentaur]

In engineering, I think, it needs flywheel and feedback to make it work,

Steam engines used to use a 'governor' to control the power delivered. This worked by adjusting the steam admitted. In the case of the spring problem, a similar centrifugal mechanism could control the angle of a cam (as shown above).

But there is a long history of clockwork power regulation. High accuracy clocks (chronometers) keep their accuracy by making sure the balance wheel has the same amplitude of oscillation all the time. The design of the escapement does a good job of amplitude regulation by appropriately shaped and angled pallets and escape wheel faces. Also the Fusee movement uses a shaped barrel (effectively a cam) with a light chain wrapped round it to compensate for the reduction in spring torque as it runs down.

If this query is for a serious practical application then the OP should read horology books. It's been studied for hundreds of years and really gets the best out of mechanical timing mechanisms. (Shame about the quartz crystal).

 

[tech99]

If you consider the electrical analogy, a spring is equivalent to a capacitor, the mass of the system is equivalent to inductance and the friction (or device against which work will be done) equivalent to resistance. The spring compliance plus a mass create a resonant system. If you extract power from it using a friction-like device, then we can adjust things so the spring delivers its energy but does not rebound. In other words, the system is critically damped.

 

[sophiecentaur]

If you consider the electrical analogy, a spring is equivalent to a capacitor, the mass of the system is equivalent to inductance and the friction (or device against which work will be done) equivalent to resistance. The spring compliance plus a mass create a resonant system. If you extract power from it using a friction-like device, then we can adjust things so the spring delivers its energy but does not rebound. In other words, the system is critically damped.

The requirement is constant Power, though. I don't think critical damping dissipates constant power, does it? Also, there are many 'easy' ways if you involve resistive losses. A cam-based system is not inherently lossy - it is a lever (aka 'transformer' ) which loses 'no' energy.

 

[jack action]

Steam engines used to use a 'governor' to control the power delivered.

A governor regulates the speed, not the power.

 

[sophiecentaur]

I acknowledged that in my post but Power can be controlled, using speed as a feedback signal. As I said, the torque can be measured and controlled. Speed times torque gives the Power. The task is more straightforward in an electrical context but a clock mechanism achieves what’s been asked for; the hands are driven at constant speed which means constant power.

 

[cairoliu]

Steam engines used to use a 'governor' to control the power delivered. This worked by adjusting the steam admitted. In the case of the spring problem, a similar centrifugal mechanism could control the angle of a cam (as shown above).

But there is a long history of clockwork power regulation. High accuracy clocks (chronometers) keep their accuracy by making sure the balance wheel has the same amplitude of oscillation all the time. The design of the escapement does a good job of amplitude regulation by appropriately shaped and angled pallets and escape wheel faces. Also the Fusee movement uses a shaped barrel (effectively a cam) with a light chain wrapped round it to compensate for the reduction in spring torque as it runs down.

If this query is for a serious practical application then the OP should read horology books. It's been studied for hundreds of years and really gets the best out of mechanical timing mechanisms. (Shame about the quartz crystal).

Classical clockwork mechanisms use delicate mechanic parts to execute complicated feedback governing algorithm.
However modem CPU based algorithms can eliminate exotic mechanisms, impart work-suckers aka loads with intelligence in order to realize designer's purpose.
Mechansim or hardware-driven algorithms have limited capacity, e.g. +, -, <, > as usual, in contrast, CPU-based algorithms can solve whatever complexity.
If there were CPUs centuries ago, full mechanism clockpieces would have no chance to be invented, but quartz watches appeared first.
Until today, nobody realizes Carnot-cycle by full mechansim, forever impossible? but who knows, maybe someday in future, CPU-controlled 4 perfect cycles of Carnot heat engine could come true.

 

[jack action]

but a clock mechanism achieves what’s been asked for; the hands are driven at constant speed which means constant power.

In the case at hand, the force constantly varies; that is the initial restriction. Therefore the speed must also vary - in a controlled manner - to get constant power. A mechanism that can keep a constant speed is thus useless.

 

[sophiecentaur]

In the case at hand, the force constantly varies; that is the initial restriction. Therefore the speed must also vary - in a controlled manner - to get constant power. A mechanism that can keep a constant speed is thus useless.

A suitable low pass filter would be a trivial addition - it's called a flywheel with a flexible drive.

 

[cairoliu]

In the case at hand, the force constantly varies; that is the initial restriction. Therefore the speed must also vary - in a controlled manner - to get constant power. A mechanism that can keep a constant speed is thus useless.

Hi, I sincerely invite honorable Jack to fix my new challenge:

What speed v(t) enables constant power by solenoid pull? ​