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

• cairoliu
In summary: The escapement is the key and then it's a matter of getting the torque to vary in a way that compensates for the variation in spring force.In summary, the conversation discusses the concept of linear spring force and its relationship to displacement and power. The question asks if there is a mathematical function that can produce constant power, and the responses propose different methods such as using a centrifugal mechanism or applying principles from clockwork power regulation. Ultimately, the topic is a complex and well-studied one, with various engineering solutions available.
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?

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.

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

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:

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.

jack action said:
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:

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.

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 said:
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.
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!

jack action
Lnewqban said:
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.

cairoliu said:
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).

anorlunda and Lnewqban
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.

tech99 said:
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
sophiecentaur said:
Steam engines used to use a 'governor' to control the power delivered.
A governor regulates the speed, not the power.

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.

sophiecentaur said:
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.

sophiecentaur said:
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.

jack action said:
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.

jack action said:
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 is the equation for calculating the speed v(t) that enables constant power by releasing a compressed spring?

The equation is v(t) = √(2P/m(1-cos(θ))), where P is the power, m is the mass of the object attached to the spring, and θ is the angle at which the spring is released.

## How does the speed v(t) affect the power generated by releasing a compressed spring?

The speed v(t) directly affects the power generated by releasing a compressed spring. The faster the speed, the more power is generated. However, if the speed is too high, the spring may become damaged and the power output may decrease.

## What factors can affect the speed v(t) needed for constant power from a compressed spring?

The speed v(t) needed for constant power from a compressed spring can be affected by various factors such as the strength and elasticity of the spring, the mass of the object attached to the spring, and the angle at which the spring is released.

## Can the speed v(t) that enables constant power from a compressed spring be calculated using only the power and mass?

No, the speed v(t) also depends on the angle at which the spring is released. Therefore, the angle must be known or measured in order to accurately calculate the speed v(t).

## What is the relationship between the speed v(t) and the power generated by releasing a compressed spring?

The relationship between the speed v(t) and the power generated by releasing a compressed spring is directly proportional. This means that as the speed increases, the power also increases. However, there may be a maximum speed at which the power output plateaus or decreases due to factors such as the strength and elasticity of the spring.

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