Valve math

[markdem]

Does anyone have a mathmatic calculation for how poppet valves interact with other parts?

I understand the basic principles, but every time i make a new valve/hammer/spring combo, im just taking guesses and i end up having to re-make parts because i need to make small adjustments

Could the equation be dumbed down to something as simple as: hammer force - valve force = dwell time?

I understand there are variables such as airflow restrictions but a basic calculation would get me started in the right direction

[Siress]

the whole thing is a sequence of differential equations... I will be shocked if there's a simplification that's any better than your gut.

[tyronejk]

Can you expand on this? At least give some starting points (e.g. approaching this from an energy conservation standpoint, or looking at kinematics in freq domain, etc)? I think I may have been able to noodle this out in college, but it's been far too long since I did any heavy math.

[Siress]

This is one of those 'if you wish to make an apple pie from scratch, you must first create the universe' conversations. Awareness of complex subjects and understanding them are very different things. Freq. domain is largely irrelevant except as it may apply to recovery of the springs. Kinematics will only get you close enough to determine boundary conditions under ideal scenarios - not reality. e.g. it takes at least 41psi to propel a paintball out of an 8 inch control bore. To bridge the gap, you have to get into dynamics. That's where most engineers will check out. Those with grit and heavy background in mathematics will wade into it at varying depths. Most of those will drown in conditional statements they don't know how to contend with all at once. It's fun at first...but it's time consuming, and that becomes a punishment after a while. Modern methods utilize software to construct and compute the problem. I would recommend ANSYS for this particular problem, but I keep wanting COMSOL to be the go-to option.

[Trbo323]

Yeah I think the effort to find all the different factors may be more work than swapping springs

Just off the top of my head

Main spring Force
Valve spring Force
Air pressure acting to close the valve
Weight of hammer
Hammer friction in lower tube
Poppet size

Finding most of those wouldn't be hard but figuring out an equation that would give you reliable results? I certainly wouldn't know where to start

Sent from my motorola edge 5G UW (2021) using Tapatalk

[Alexndl]

Its been a while that I looked at those formula but I would start with the air expansion from the valve and the barrel to propel the paintball (the fun math pats its the volume vary over time). Chamber volume, pressure initial speed of the paintball, final speed of the paintball, friction and some pressure inefficiency.

Then you can calculate the time you want the valve to open wirh with an impact calculation to optimise the efficiency. Would need to use the mass of the hammer, speed of the hammer at contact with the valve pin which can be calculated with main spring rate, compression distance, travel distance between is original location and the valve, minus the friction of the hammer inside the body. On the other side you will have the force from the valve pin. Which will be valve spring rate, maximum compression distance and the compress air force on the valve head (pressure and surface area but only the cross section perpendicular to the valve movement). The issue with this the pneumatic force on the valve is the pressure will drop quickly as soon as the valve will open so not sure how to handle this one.

Just thinking I would need to write those formula make my nose bleed... I hated that course in university.

I would guess finite element calculator would make this a lot easier.

[redfireant3]

Might be able to load the values into ChatGPT and get a value(maybe it'll be correct?). Or to use different variables ask Copilot in Edge to make a csv or Powershell script to make an Excel file or something for you so you change the variables.

[downfall]

Does this help?

[markdem]

Very similar. Im sure some of it can be transferred over. Good find!

[downfall]

I've been using Claude.ai ...

Example response:

To derive a formula for the mechanical dwell time on a spool valve based on a specific spring force and hammer force, we need to consider the dynamics of the spool valve system and apply Newton's second law of motion.

Let's define the following variables:

- m = mass of the spool
- x = displacement of the spool from its equilibrium position
- k = spring constant of the return spring
- F_h = force exerted by the hammer
- F_f = frictional force opposing the spool motion
- t_dwell = mechanical dwell time

Assumptions:
1. The spool moves in a frictionless environment (F_f = 0).
2. The hammer force (F_h) is constant during the dwell time.
3. The spring force follows Hooke's law (F_s = -kx).

According to Newton's second law of motion, the sum of forces acting on the spool is equal to the mass times acceleration:

∑F = m(d^2x/dt^2)

Substituting the forces acting on the spool:

F_h - kx = m(d^2x/dt^2)

Rearranging the equation:

m(d^2x/dt^2) + kx = F_h

This is a second-order differential equation, and its solution depends on the initial conditions of the spool's position and velocity.

Let's assume that the spool starts from rest at its equilibrium position (x = 0, dx/dt = 0) at time t = 0.

Integrating the equation twice with respect to time, we get:

dx/dt = (F_h/k)(1 - cos(ωt))
x = (F_h/k^2)(ωt - sin(ωt))

where ω = √(k/m) is the natural frequency of the spool-spring system.

The mechanical dwell time (t_dwell) is the time it takes for the spool to reach its maximum displacement (x_max) before reversing its motion. At the maximum displacement, the velocity (dx/dt) is zero.

Setting dx/dt = 0 and solving for t_dwell, we get:

t_dwell = (π/ω) = π√(m/k)

Substituting the expression for ω, we get:

t_dwell = π√(m/k)

This formula gives the mechanical dwell time (t_dwell) on a spool valve based on the mass of the spool (m) and the spring constant of the return spring (k).

Note that this formula assumes a constant hammer force (F_h) and neglects friction. In reality, the hammer force may vary over time, and friction forces may also affect the dwell time. Additionally, if the spool encounters physical stops or other nonlinear effects, this linear model may not accurately represent the system's behavior.

Poppet valve should be the same

No, the dwell time (mechanical delay time) of a spool valve is not entirely independent of the force of the hammer acting upon it. The hammer force does play a role in determining the dwell time, although it is not directly accounted for in the simplified formula we derived earlier.

In the previous derivation, we assumed a constant hammer force (F_h) acting on the spool during the dwell time. This assumption led to the following equation of motion:

m(d^2x/dt^2) + kx = F_h

Where:
- m is the mass of the spool
- k is the spring constant of the return spring
- x is the displacement of the spool
- F_h is the constant hammer force

The solution to this equation gave us the dwell time formula:

t_dwell = π√(m/k)

However, in reality, the hammer force is not constant but varies over time. The force exerted by the hammer is typically a pulse or an impulse, which means that the force rises rapidly to a peak value and then decays quickly.

The magnitude and duration of this hammer force pulse will affect the initial acceleration of the spool and, consequently, the dwell time. A stronger hammer force (higher peak force or longer duration) will impart more initial acceleration to the spool, potentially reducing the dwell time.

Additionally, if the hammer force is large enough to overcome the spring force and cause the spool to impact against physical stops or buffer mechanisms, the dwell time calculation becomes more complex, as it needs to account for the impact dynamics and energy dissipation.

Therefore, while the simplified formula we derived provides a good approximation for the dwell time based on the spool mass and spring constant, it does not fully capture the effect of the hammer force. To more accurately model the dwell time, you would need to consider the specific hammer force profile and incorporate it into the equations of motion, potentially using numerical simulations or empirical testing.

In summary, the dwell time of a spool valve is influenced by both the spool-spring dynamics and the characteristics of the hammer force acting on the spool.

I got excited but this is the more appropriate response?

To derive a more accurate expression for the dwell time of a spool valve, we need to consider the separate phases of motion and account for the varying forces acting on the spool during each phase. Here's a step-by-step derivation:

Acceleration phase:
During this phase, the spool is accelerated from rest by the hammer force (F_h). The equation of motion is:

F_h - kx = ma

Integrating twice with respect to time, and assuming initial conditions of x(0) = 0 and v(0) = 0, we get:

v = (F_h/m)t - (k/m)x
x = (F_h/2m)t^2 - (k/6m)t^3

Let t_a be the time at the end of the acceleration phase, when the spool reaches its maximum velocity (v_max).
Substituting x = x_max and v = v_max into the above equations, and solving for t_a, we get:

t_a = (m/k)[(3v_max/x_max)^(1/2) - (v_max/x_max)^(1/3)]

Constant velocity phase:
During this phase, the spool travels at a constant velocity (v_max) for a time t_c.
The displacement during this phase is:

x_c = v_max * t_c

Deceleration phase:
During this phase, the spool decelerates due to the spring force until it comes to rest.
The equation of motion is:

-kx = ma

Integrating twice with respect to time, and using the initial conditions v(t_a + t_c) = v_max and x(t_a + t_c) = x_max + x_c, we get:

v = v_max - (k/m)(x - x_max - x_c)
x = x_max + x_c + v_max*t - (k/2m)t^2

Let t_d be the time at the end of the deceleration phase, when the spool comes to rest (v = 0).
Substituting v = 0 into the velocity equation and solving for t_d, we get:

t_d = (m/k)v_max

The total dwell time (t_dwell) is the sum of the times for all three phases:

t_dwell = t_a + t_c + t_d

Substituting the expressions for t_a, t_c, and t_d, we get:

t_dwell = (m/k)[(3v_max/x_max)^(1/2) - (v_max/x_max)^(1/3)] + (x_max/v_max) + 2(m/k)v_max

This expression for the dwell time accounts for the separate phases of motion and the varying forces acting on the spool during each phase. It depends on the following parameters:

- m: mass of the spool
- k: spring constant of the return spring
- F_h: hammer force (used to calculate v_max and x_max)
- v_max: maximum velocity of the spool
- x_max: maximum displacement of the spool

Note that this derivation assumes a constant hammer force (F_h), neglects friction, and considers a linear spring force. In reality, the hammer force may vary over time, and other non-linear effects may need to be accounted for, leading to even more complex expressions for the dwell time.

[markdem]

It would be easier to make a new valve instead of figure any of that out

[Siress]

and that, dear reader, is engineering in a nutshell.