How fast Coil Springs are? Greatest mystery today?

[Wes Tausend]

Don't know if this helps any but as a machinist I have encountered a condition called "float" many times in race engines. The compressive strength of the spring steel has to be increased ( stouter springs installed) in high rpm racing applications because the valves can't close fast enough for the compression stroke. Near 8500 rpm with standard springs and 10000+ with high performance. So with the compressed installed length - maximum compressed installed length rpm factored and graphed and float being the indication fail @ maximum speed you could set up a mechanical experiment to check the "speed of a spring". Heat is an issue here too.

I hope Darp (OP) does not mind that we explore this.

I kind of wondered if one couldn't empirically determine spring speed by resonance. Perhaps Darp has done this. A spring can't logically be any faster than its ability to return from a tension. The idea would be to drive the spring from one end with transducer and calibrated audio generator and observe when it reached maximum resonance, or "bounce" on the other end. Some lesser resonances would appear at harmonic frequencies, but the true primary resonance would be observed to be the most vigorous. Ideally one would also have a frequency analyser if great accuracy was needed.

In the practical cases of engine valve float, one could assemble the proposed valve train and poppet valve which should cause the effective resonant frequency to lower because of greater inertia. It could not be allowed to lower any more than the maximum design rpm of the engine which is within keeping of using stiffer springs.

This same principle should also apply to a vehicle suspension spring when the unsprung mass (suspension arms, wheel etc) is attached. The idea here is to return the wheel for continuous traction and could be best theoretically tuned to the known average bump frequency on a moto-cross track for instance.

Thanks,
Wes
...

 

[jack action]

Thank you Chet for taking the time to write the detailed equations. AlephZero might called it the "hard way", but I have to say that good old "F=ma" help me see the solution easier as I'm not familiar with the concept of wave equation.

Incidentally, a simpler way of solving this problem is to use conservation of energy: the kinetic energy of the spring after separating from the second constraint (at x = 0) must be equal to the initial stored elastic energy in the spring. Assuming that all parts of the spring are moving at the same velocity after total separation (as indicated by my detailed analysis in #16), we have:

\frac{1}{2}mv^2=\frac{1}{2}ku_0^2

This again gives:

v=u_0\sqrt{\frac{k}{m}}

This relates to what I told you in post #17 as I recognized the equation. I tought you did a very simplistic evaluation.

This is what stumble me. This is the exact way to find the velocity of a mass attached to a single massless spring. In my post I was saying "Happy coïncidence ?" as I was under the impression that if spring mass was involved, it wasn't simply added to the end mass. IIRC, with problems concerning car suspensions, when time came to determine if a suspension component (i.e. spring, shock, a-arm) was supposed to be considered "sprung mass" or "unsprung mass", the rule of thumb was to divide the mass 1/3-2/3 between sprung and unsprung mass (or vice-versa). I never bother to do detailed calculations, but it seemed logic that it couldn't be simply put at one end of the spring (or split in half between the two masses in this case).

What is also puzzling me is that if we refer to the email in the OP, they are 2 equations: one that is basically yours and another similar one that reads:

For springs with a mass ratio (P/Ws) >4

v= F √ ( kg/P+1/3Ws)

I wonder what happens with those lighter springs that they don't "obey" the same rule ? Plus, the units don't fit, so it must be empirical, and there is that familiar 1/3 of spring mass that appears from I don't know where.

 

[Chestermiller]

Thank you Chet for taking the time to write the detailed equations. AlephZero might called it the "hard way", but I have to say that good old "F=ma" help me see the solution easier as I'm not familiar with the concept of wave equation.



This relates to what I told you in post #17 as I recognized the equation. I tought you did a very simplistic evaluation.

This is what stumble me. This is the exact way to find the velocity of a mass attached to a single massless spring. In my post I was saying "Happy coïncidence ?" as I was under the impression that if spring mass was involved, it wasn't simply added to the end mass. IIRC, with problems concerning car suspensions, when time came to determine if a suspension component (i.e. spring, shock, a-arm) was supposed to be considered "sprung mass" or "unsprung mass", the rule of thumb was to divide the mass 1/3-2/3 between sprung and unsprung mass (or vice-versa). I never bother to do detailed calculations, but it seemed logic that it couldn't be simply put at one end of the spring (or split in half between the two masses in this case).

What is also puzzling me is that if we refer to the email in the OP, they are 2 equations: one that is basically yours and another similar one that reads:



I wonder what happens with those lighter springs that they don't "obey" the same rule ? Plus, the units don't fit, so it must be empirical, and there is that familiar 1/3 of spring mass that appears from I don't know where.

Hi Jack,

I can't really comment on the equations Darp posted from the "Institute", because they don't provide any details. All I can say is that I'm totally confident about the formulation and solution that I obtained. As far as my formulation is concerned, it is fully consistent with the multi-spring/mass model that you proposed, in the limit of a continuous distribution of springs and masses. And the resulting partial differential equation was straightforward to solve for the specified initial and boundary conditions. Further, the final result for the release velocity was consistent with the result obtained using conservation of energy.

As an aside, I might mention that this development is almost identical to what one can use to elucidate the details of an elastic collision between two cylindrical masses colliding head-on.

Chet

 

[Q_Goest]

Perhaps looking at it from the perspective of a spring's natural frequency would help.

A spring, mass system oscilates at a natural frequency equal to √(k/m). As you decrease m, the frequency increases until you have only the mass of the spring itself operating. That frequency is the natural frequency of the spring as given by:

This is from efunda which provides the derivation.

This doesn't give velocity of any portion of the spring directly, but the motion of any point on the spring follows a sinusoidal path.

The highest velocity of any point on the spring then is at the free end of the spring when it passes through that point it would be at when the spring is stationary. That velocity is a function of how far the spring is initially compressed or the amplitude of the movement. The higher the initial compression, the higher the velocity as it passes through its unstressed shape.

I think velocity should be relatively straightforward to derive from this.

 

[Darp]

Great resource

I empirically solved the question, but am so impressed with the knowledge here. I am not where close to being able to do the math you folks have. And is case you missed it, the the empirical answer is about 50 mph, springs are slow, because they accelerate them own large mass.

 

[Darp]

Perhaps looking at it from the perspective of a spring's natural frequency would help.

A spring, mass system oscilates at a natural frequency equal to √(k/m). As you decrease m, the frequency increases until you have only the mass of the spring itself operating. That frequency is the natural frequency of the spring as given by:

This is from efunda which provides the derivation.

This doesn't give velocity of any portion of the spring directly, but the motion of any point on the spring follows a sinusoidal path.

The highest velocity of any point on the spring then is at the free end of the spring when it passes through that point it would be at when the spring is stationary. That velocity is a function of how far the spring is initially compressed or the amplitude of the movement. The higher the initial compression, the higher the velocity as it passes through its unstressed shape.

I think velocity should be relatively straightforward to derive from this.

I can not do the math, but am very good at feel or whatever you want to call it. The max velocity would come before relaxed state, JMHO Because the last 20% of travel is at much less accelerating force, yet is pushing same mass same distance as the 50-70% extension. In fact maybe the total mass/distance is more.

Just my guess the max velocity would be about 80% of release distance.

 

[Q_Goest]

Hi Lynne, The efunda site I took the above equation from indicates this is the "first natural frequency" for a spring with no mass attached and I've described what I understand that shape to be. This seems to agree with the roymech.co.uk descrition here, so I'm pretty sure we can describe a spring with distributed mass using this equation as I've suggested. But this is getting away from my area of experise. I suspect you could get other modes developing (ie: second, third, etc... natural frequencies) so this assumes the spring will oscillate at its first natural frequency. I assume that's a safe assumption...

To go from frequency to determining position and velocity however is relatively straightforward. First, calculate the angular frequency ω (rad/s), by multiplying frequency times 2π. The position then is basically:
x=A*sin(ωt)
and velocity is:
v=A*cos(ωt)
where A is amplitude, or how far you compress or stretch the spring from its neutral location and t is time in seconds.

That's about it. If you have a specific spring in mind and want to see what this approach predicts for maximum velocity and if you don't feel comfortable doing the math, just provide:

• wire diameter
• nominal coil diameter (average of ID and OD)
• number of coils
• the generic type of material (ie: steel (includes music wire for example), stainless steel, copper alloy, etc...)
• How far the spring is compressed from its neutral position before being released

 

[Darp]

Don't know if this helps any but as a machinist I have encountered a condition called "float" many times in race engines. The compressive strength of the spring steel has to be increased ( stouter springs installed) in high rpm racing applications because the valves can't close fast enough for the compression stroke. Near 8500 rpm with standard springs and 10000+ with high performance. So with the compressed installed length - maximum compressed installed length rpm factored and graphed and float being the indication fail @ maximum speed you could set up a mechanical experiment to check the "speed of a spring". Heat is an issue here too.

Yes, That was first thing I did, and they came out slow, I excused it to being it has to push that heavy valve. Did all the math (lift and duration) and it came out about 40 fps. After that came here. Then did spring empirical speed test several ways, springs are just slow.

Thanks.

 

[Darp]

Hi Lynne, The efunda site I took the above equation from indicates this is the "first natural frequency" for a spring with no mass attached and I've described what I understand that shape to be. This seems to agree with the roymech.co.uk descrition here, so I'm pretty sure we can describe a spring with distributed mass using this equation as I've suggested. But this is getting away from my area of experise. I suspect you could get other modes developing (ie: second, third, etc... natural frequencies) so this assumes the spring will oscillate at its first natural frequency. I assume that's a safe assumption...

To go from frequency to determining position and velocity however is relatively straightforward. First, calculate the angular frequency ω (rad/s), by multiplying frequency times 2π. The position then is basically:
x=A*sin(ωt)
and velocity is:
v=A*cos(ωt)
where A is amplitude, or how far you compress or stretch the spring from its neutral location and t is time in seconds.

That's about it. If you have a specific spring in mind and want to see what this approach predicts for maximum velocity and if you don't feel comfortable doing the math, just provide:

• wire diameter
• nominal coil diameter (average of ID and OD)
• number of coils
• the generic type of material (ie: steel (includes music wire for example), stainless steel, copper alloy, etc...)
• How far the spring is compressed from its neutral position before being released

Thanks, I am already on work another direction, so no longer needed, but appreciate it.

 

[hispanic panic]

I just want to chime in here because I'm genuinely interested in release characteristics of coil springs. I didn't read a lot of what was typed here as I'm pretty lazy, but one thing that i didn't see mentioned was the damping of the spring material itself. Simple fact of the matter is that to correctly model what is happening at the speeds a spring releases, you MUST model the damping of the material itself into the equations. Different types of steel will damp itself differently upon expansion. The company Swift prides itself on a special kind of steel that is stiffer than a typical coil spring, allowing them to get away with less coils for a given rate. Swift also touts a much more "responsive" spring I'm assuming due to its lower mass. The material itself could also have a lower damping rate in comparison, but I'm not sure this difference can be measured easily.

For a car application, i emplore you to google "shock dyno". A shock dyno essentially moves a shock through various velocities and measures the damping force in rebound and compression. In real aplications, the rebound speed of a spring will be damped by the shock absorber as well, so overall response will be damped even further. Shock dyno's start at 0 in/sec and go all the way to 8-14 in/sec. Most people stop at 8, but motorsports and OEM applications look at speeds double that rate as high end shock absorbers will have blow off mechanisms to dramatically decrease damping force at extremely high velocities. This is where spring response is important, and where a spring like swift would shine.

Overall, i don't think we're taking the right approach to modeling this situation. I feel it will be more of a spring/mass equation with a damper and spring acting in parallel, with both the spring and damper having a separate force and damping component for both (dampers are usually pressurized, which would provide a variable spring force).