How to increase Lego Technic Shock Absorber stiffness

[Atr]

Have you ever needed Small Technic Shock Absorber 6.5 L that have harder springs than grey, but softer than yellow ones? All you have to do is put a single Technic Half Bush on a piston rod like on the video below.

I came up with this idea last year when I was building my Jeep Willys.

[Jurgen Krooshoop]

Nide idea, thanks for sharing.

[Milan]

Great. Very usable idea!

[DLuders]

Atr, I will try the toothed 1/2 bushing mod on my next creation. The video showed three different colors of Lego Technic Shock Absorbers, but there are more colors. Below is a chart that shows the decreasing spring resistance (in grams) of standard shocks. I can't remember where I got it from on the internet. As you noted, yellow is the stiffest (at 1300 grams), Dark Grey is next at 600 grams, then Light Grey at 520 grams and and Dark Turquoise is the weakest at 430 grams. I wonder if placing a toothed 1/2 bushing on a Light Grey shock will make it more stiff than a Dark Grey shock or not....

For the RED 6.5L shock absorber, Bricklink lists Part 731c01 as "Technic, Shock Absorber 6.5L, Complete Assembly (Spring Undetermined)", so I don't know what spring strength it has. The Dark Turquoise 6.5L shock is 731c06, "Technic, Shock Absorber 6.5L, Complete Assembly (Soft Spring)", but apparently it is not as "soft" as the Light Grey shock. The 731c05 6.5L "Normal Spring" shock comes in the "Known Colors" of Dark Grey, Light Grey, and Red. See http://www.bricklink.com/catalogList.asp?q=technic+shock for all of the possibilities.

Edited May 18, 2010 by dluders

[Atr]

In my opinion red's and light grey's shocks stiffness are quite the same. I think that half-bush-mod on light gray shock makes them about 30% stiffer. This is a big approximation but I think they are more stiff than plain dark grey shocks. If you have them both try them and post a result.

The tests (and photo) you have posted are probably done by Philo if I remember correct.

[Blakbird]

In my opinion red's and light grey's shocks stiffness are quite the same. I think that half-bush-mod on light gray shock makes them about 30% stiffer. This is a big approximation but I think they are more stiff than plain dark grey shocks. If you have them both try them and post a result.

Strictly speaking the "stiffness" or spring rate of the shock is the same whether you put a bushing on it or not. The spring rate (typically denoted with a "k") is a constant for most springs and is the amount of linear (or rotational) motion per unit force. In imperial units this is typically pounds/inch or Newtons/centimeter in SI. Adding a bushing increases the preload on the spring by making it start in a more compressed state and therefore with a higher initial load. The additional force to make it compress a certain amount remains the same.

With all that being said, this is still a great idea! It will have the effect of providing more support for heavy models. Shock absorbers for R/C cars and trucks almost all contain a similar mechanism for changing the length.

Now if only we can find a way to make a stiff spring softer.....

[johanby]

Strictly speaking the "stiffness" or spring rate of the shock is the same whether you put a bushing on it or not. The spring rate (typically denoted with a "k") is a constant for most springs and is the amount of linear (or rotational) motion per unit force. In imperial units this is typically pounds/inch or Newtons/centimeter in SI. Adding a bushing increases the preload on the spring by making it start in a more compressed state and therefore with a higher initial load. The additional force to make it compress a certain amount remains the same.

With all that being said, this is still a great idea! It will have the effect of providing more support for heavy models. Shock absorbers for R/C cars and trucks almost all contain a similar mechanism for changing the length.

Now if only we can find a way to make a stiff spring softer.....

Yes and no. The total force required to fully compress the spring would be the same with or without the bushing (I guess the masses 600g, 1300g, etc, indicated in the picture above was the the corresponding force to fully compress the spring). On the other hand, the total travel distance decreases when inserting the bushing. Hence the stiffness (i.e. the ratio between applied load and compressed distance) will increase. Therefore the spring will be stiffer, but it cannot carry a higher load until fully compressed.

[Atr]

Thanks Blakbird for infos.

Yes and no. The total force required to fully compress the spring would be the same with or without the bushing (I guess the masses 600g, 1300g, etc, indicated in the picture above was the the corresponding force to fully compress the spring). On the other hand, the total travel distance decreases when inserting the bushing. Hence the stiffness (i.e. the ratio between applied load and compressed distance) will increase. Therefore the spring will be stiffer, but it cannot carry a higher load until fully compressed.

Note that the plain shock doesn't compress the spring fully under full load. Toothed Half Bush doesn't decrease travel distance - it makes the spring compress fully.

[Blakbird]

Yes and no. The total force required to fully compress the spring would be the same with or without the bushing (I guess the masses 600g, 1300g, etc, indicated in the picture above was the the corresponding force to fully compress the spring). On the other hand, the total travel distance decreases when inserting the bushing. Hence the stiffness (i.e. the ratio between applied load and compressed distance) will increase. Therefore the spring will be stiffer, but it cannot carry a higher load until fully compressed.

I disagree. Spring rate cannot be changed, it is inherent in the properties of the spring. It is true that the totally compressed force remains the same. The bushing changes both the starting length and the starting load but not the rate. The additional load per unit inch of deflection is the same with or without the bushing and must always be based on spring physics.

Here is an example.

Case 1: Unmodified sample shock absorber.

Free Length: 3 cm				  Free Load:  0 g
Compressed Length:  1 cm		   Compressed Load:  1000 g
Delta Length (stroke):  2 cm	   Delta Load:  1000 g
Spring Rate:  1000 g / 2 cm = 500 g/cm

Case 2: Modified sample shock absorber (0.25cm bush).

Free Length: 2.75 cm			   Free Load:  125 g
Compressed Length:  1 cm		   Compressed Load:  1000 g
Delta Length (stroke):  1.75 cm	Delta Load:  875 g
Spring Rate:  875 g / 1.75 cm = 500 g/cm

In either case we have a 500 g/cm spring, but in the second case we need to apply at least 125g of load before the spring starts moving which gives the impression of more stiffness. Another way to put it would be that the spring rate (stiffness) is a function of the spring, but the spring loading is a function of installation (free and compressed length). The modifications proposed change the installation but not the spring.

Edited May 18, 2010 by Blakbird

[johanby]

In either case we have a 500 g/cm spring, but in the second case we need to apply at least 125g of load before the spring starts moving which gives the impression of more stiffness. Another way to put it would be that the spring rate (stiffness) is a function of the spring, but the spring loading is a function of installation (free and compressed length). The modifications proposed change the installation but not the spring.

Well, this is where I disagree. All materials (OK, most materials) are linear elastic. That means that an applied stress (i.e applied force for constant cross section) will yield a (compressive or tensile) strain (and thus elongation/compression) which is proportional to the applied stress (the so called Hooke's law). To precompress the spring will not have the effect that you need to reach some limit stress/force until a strain/compression will appear (since that would mean that the material was not linearly elastic). An applied stress/force on a pre-compressed spring will be compressed, however not as much as on the spring without precompression. This means that the spring (assembly) is effectively stiffer.

It is easy to test by hand.

[sverre]

Johanby, you are wrong. Blackbird is right.

For in depth knowledge read this article:

http://www.sportrider.com/tech/146_9510_tech/index.html

[Blakbird]

Well, this is where I disagree. All materials (OK, most materials) are linear elastic. That means that an applied stress (i.e applied force for constant cross section) will yield a (compressive or tensile) strain (and thus elongation/compression) which is proportional to the applied stress (the so called Hooke's law). To precompress the spring will not have the effect that you need to reach some limit stress/force until a strain/compression will appear (since that would mean that the material was not linearly elastic). An applied stress/force on a pre-compressed spring will be compressed, however not as much as on the spring without precompression. This means that the spring (assembly) is effectively stiffer.

This is fascinating thought exercise (and incidentally I design and analyze aircraft springs and mechanical systems for a living). You're really close with your thought process.

You are correct that most materials (at least ductile metals) are linear elastic at least up until their proportional limit stress at which point they exhibit plasticity and will not return to their original shape but rather exhibit hysteresis. When speaking of practical springs, we're never going to be loading them near the proportional limit, so we'll always consider them linearly elastic. In other words, stress and strain are proportional. A spring (or beam) with double the load will have double the deflection. This is in fact exactly the reason that the spring rate cannot change as a function of initial length. The slope of the stress/strain curve (Young's Modulus) is a constant in the linear elastic range and no amount of preload will change that unless you yield the spring (which ruins it).

The issue here comes in the fact that you are saying it is impossible to add external force to the system and NOT produce some strain. This is a correct statement, but it does not therefore follow that the spring compresses even at an applied load below the preload threshold. The reason for this is hard to explain without a free body diagram, but I'll try. All static systems must be in equilibrium. When a preloaded shock absorber is at it's free length, the spring is under a certain amount of compression. The body of the shock must be in an equal and opposite state of tension. In this case, the yellow cylinder and the black piston are being forced apart by the same force that produces compression in the spring. What is preventing them from separating? The internal stop. The little "forks" inside the rod are reacting against the slots in the cylinder. This load path is carrying the 125g of tension in my example. The reason that you can apply an external load without compressing the spring is that the load is not being carried by the spring. The first 125g of external load goes into relieving the existing tension that is being carried by the stop. Only when this tension has been nullified can the spring again become part of the load path of external compression.

Incidentally, there IS strain in the system as this external load is being applied, but it is a small tensile strain in the piston and cylinder which is being relieved. The piston and cylinder are so stiff when compared with the spring that this strain is not visually noticeable. This is why you don't see it when you just try to solve the problem while holding one in your hand. Hooke's law is not being violated.

Upon further reflection, I decided to include a free body diagram. The purple force (S) represents the spring. The blue force (E) is external. The red force (P) is the amount of preload being carried by the stop. Now consider equilibrium.

• When the external load is zero (unloaded), the spring force is balanced by the preloaded stop. S=P
  
• When a significant load (greater than the preload) is applied externally, the spring compresses and there is no longer any load on the stop. In this case the external load is equal to the spring load. E=S
  
• However, when the external load is less than the preload, it is reacted by decreasing the preload in the stop. P=S-E The spring force stays the same but the preload goes down until the stop separates.
  

If you really want to get super technical (which I guess we are) then the preloaded piston and the spring are both strained with external load as a function of their respective spring rates (springs in parallel). However, the spring rate of the piston is so much large than the spring rate of the spring that the proportional strain is effectively all in the piston until the preload is relieved.

I've reviewed plenty of stress reports of hydraulic actuators by professional engineers who have still not managed to grasp this principle, so don't feel bad!

[DLuders]

Maybe the real answer is NOT TO LIVE WITH the limited selection of Lego springs, but to GET SOME DIFFERENT SPRINGS from the McMaster-Carr Catalog, or some Spring supplier somewhere. More coils or thicker wire --> STIFFER SPRINGS. Does anybody know where to get suitable springs that could work on 6.5L or 9.5L Technic Shock Absorbers? Check out these possibilities: http://www.mcmaster.com/#compression-and-die-springs/=75e4ev .

Edited May 19, 2010 by dluders

[johanby]

Incidentally, there IS strain in the system as this external load is being applied, but it is a small tensile strain in the piston and cylinder which is being relieved.

That is the key point! Thank you for a very good explanation, now this makes sense to me. Obviously, the precompression only moves us to another point on the force-displacement curve, with the slope/stiffness being the same. So it means that the constitutive equation instead would be

F=F, for F<F0

F-F0=kx, for F>F0

[Blakbird]

That is the key point! Thank you for a very good explanation, now this makes sense to me. Obviously, the precompression only moves us to another point on the force-displacement curve, with the slope/stiffness being the same. So it means that the constitutive equation instead would be

F=F, for F<F0

F-F0=kx, for F>F0

Woohoo! I'm always pleasantly surprised when I am able to successfully explain something complex. Your equations are essentially correct. The slope/stiffness of the system is therefore bilinear and continuous. Again, to be very specific the first equation is actually also F=kx, but in that case k is the stiffness of the piston, not the spring. In the second equation k is the stiffness of the spring.

[johanby]

So I guess the only way to change the stiffness of a spring is to change its length. A non-destructive way would be to make combinations of springs in series or/and in parallel. Two springs with stiffnesses k1 and k2 would then have an effective stiffness k1+k2 in parallel and k1*k2/(k1+k2) in series. A destructive way, but with more freedom, would be to cut the spring (gives a stiffer spring) or add extra pieces (gives a less stiff spring, however care must be taken so they not overlap (sit in parallel), effectively making a stiffer spring). On the other hand, this procedure also changes the precompression...

[Blakbird]

So I guess the only way to change the stiffness of a spring is to change its length. A non-destructive way would be to make combinations of springs in series or/and in parallel. Two springs with stiffnesses k1 and k2 would then have an effective stiffness k1+k2 in parallel and k1*k2/(k1+k2) in series. A destructive way, but with more freedom, would be to cut the spring (gives a stiffer spring) or add extra pieces (gives a less stiff spring, however care must be taken so they not overlap (sit in parallel), effectively making a stiffer spring). On the other hand, this procedure also changes the precompression...

If you ever design a spring from scratch, you find that the stiffness (spring rate) is a function of:

• The material (nearly constant 30e6 psi modulus for steels)
  
• The wire diameter
  
• The coil diameter
  
• The pitch
  

Once you have a spring made, all of these things are fixed. If you want the spring to fit over the LEGO shock, the coil diameter is also fixed, so all you have to play with is wire diameter and pitch. You can see that the different LEGO shock springs have different pitch. If you make the wire diameter too large, then the solid height is very long and it won't compress much.

[lego12345]

Use progressive springs. It worked on my real motorcycle.

http://www.ehow.com/facts_7462120_progressive-fork-springs_.html

[xsysiek]

Thanks Blakbird for infos.

Note that the plain shock doesn't compress the spring fully under full load. Toothed Half Bush doesn't decrease travel distance - it makes the spring compress fully.

If you ever need spring specially designed try this one http://www.dudaduda.pl/index.php?strona=glowna This guy could make you any spring, just give him dimensions and load. I used his springs twice and were very good produced.

Then you don't need to use this kind of method.

[Milan]

If you ever need spring specially designed try this one http://www.dudaduda.pl/index.php?strona=glowna This guy could make you any spring, just give him dimensions and load. I used his springs twice and were very good produced.

Then you don't need to use this kind of method.

Its about solving the issue using LEGO parts, not buying custom items.

After all it is personal choice, Atr showed how to make harder shocks using only LEGO parts.

But thanks for info, it may be useful for somebody.

[Stanza]

Its about solving the issue using LEGO parts, not buying custom items.

After all it is personal choice, Atr showed how to make harder shocks using only LEGO parts.

But thanks for info, it may be useful for somebody.

Another way to increase the stiffness, or to increase the travel of the shock....

Would be to change the angle of force applied to the shock.

ie

Using the same shock absorber, but applying the force "thru" a lever would either increase or decrease the travel... thus increase or decrease the amount of travel / force needed to compress the spring.

eg

Change from say the way the suspension of the 8297 works....wheel travel up thru the shock absorber..

To say the way the 8465 or 8458 "silver champion....where the wheel travel travels thru a lever then up thru the shock absorber.