Sim Racing servo ffb systems : OSW & Bodnar

Refers to the image below:



The steering torque is the green one, the Gyro reacting torque is the red one (90° torque vector).
When you apply a steering torque the reacting one is discharged trough the suspension wishbones and nothing is transferred to the steering column (well, nearly nothing...).
I believe this is what DrRipper was saying...

 

You are right. The main component of the resulting gyroscopic torque would be absorbed by the axle bearings.

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Yes that reacting tourgue is not transfered to steering column, but turning spinning wheel takes effort.

When you are turning a wheel the front and the backside of the wheel has velocity, but middle parts (upper part and lower part) doesn't have velocity. I mean velocity caused by turning the wheel. Not spinning velocity. If you consentrate on one point of the wheel the point has no velocity when it is in upper or lower part of the wheel, but when that point is in the front or back of the wheel it has velocity. The mass of that point then has been accelerated to that velocity. When the point goes back to middle parts of the wheel it's velocity is decelerated. Both of these thing requires force to happen and the force comes from steering column.

This might be very difficult to understand without pictures and my English is not too good either. Sorry for that.

Edit: These things was result of my own thinking so it can be wrong. If i think it now the decelerating part might cause a force that compensates the force needed for the accelerating part.

 

Sure. Say you wanted to pitch a helicopter forwards (i.e. nose down), you need the rotor in tail/rear position to go up and the nose position to go down. Logically you'd want to apply a force at these locations (up at the tail position and down at the nose position) to generate a net torque action that'll pitch the helicopter forwards (nose down). To generate such a net up and down force on the tail and nose respectively, you'd make the blade have more angle of attack (i.e. more lift) in the tail position and less angle of attack (i.e. less lift) in the nose position. This sounds good and all but it's wrong. Doing the above would actually result in a right roll instead of a forward pitch. Why? Because of gyroscopic effect/action. How come? Because the rotor blades of a helicopter acts as a disc and a rotating disc with mass is subject to the gyroscopic effect. Resultantly, effects occur 90 degrees later. You can basically think (in the case of pure gyroscopic effect/action) of the resultant output location taking place 90 degrees out of phase (i.e. later/delayed) from the input location.

To explain why this is so can be a long winded and difficult task especially without illustrations (or at least for me). This is the best video i know of explaining exactly how and why it occurs.

Despite at first seeming to break the laws of physics, the gyroscopic effect can be very beautifully explained by nothing more than newtons three laws of motion.

Yes, that's the gyroscopic effect in part (though a large part is actually from the trail due to caster as demonstrated here) but even so, this effect is apparent on a bike because it's in contact with the ground. Take it off the ground and it should not be felt through the steering just like with the hypothetical off-ground FWD car. Technically it would make the vehicle body (and you on it) roll but unless the wheel is spinning incredible fast, but again, you won't feel it through the steering unless your on the ground. In the air, the vehicle could roll slightly but massively dependent on wheel speed and mass of it vs entire body, etc, etc. Different story. lol.

 

Yes thank you.

Illustrations are the bomb.

 

My opinion now is that if wheel is attached rigidly like it is in car, spinning of the wheel doesn't make it heavier to turn. But if wheel can rotate around horizontal axel also, part of the force that is directed to turn the wheel around vertical axel goes in turning the wheel around horizontal axel also. That's why it would feel heavier to turn the wheel.

So if you take front wheel of a bike of the ground the gyroscopic effect still makes turning the wheel heavier, but if bike is attached to something so that it can't lean the steering works like in car (no resistance caused by gyroscopic effect)

 

Shouldn't do. Imagine being in space and being strapped to the seat of a bike. No matter how fast or slow the front steering wheel spins, it will feel the same in resistance. The different however is that if the front wheel is spinning, if you turn it it will cause the entire body of mass (i.e. you + the bike) to roll. And the faster the spin of the front wheel the faster you will roll for the same steering input.

Now imagine the same but on earth and with the wheels on the ground (thus your in motion). Turning the steering wheel will result in the bike rolling in the opposite direction and you thus a shifting in the center of gravity in line with the trail from the caster of the front steering wheel which will produce a self-aligning/restoring torque which is what you feel in the steering wheel as the steering wheel getting "heavier".

 

From what I have understood turning the wheel right or left wouldn't have an overall torque effect on steering direction but on the camber direction, trying to increase inner wheel camber and decrease outer wheel camber.

Tire load, which is eccentric however does produce a torque around car longitudinal axis this produces opposite effects in each front wheel in convergence/divergence direction. when tire loads get unbalanced (which happens in turns or bumps) they produce the normal FFB we all know: the right wheel pushes left and the left wheel pushes right. Camber and caster settings provide a significant difference on this effect since they alter the contact patch and hence the eccentricity of tire load.

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What do you mean by that? By reading your post i think our opinion about this are similar, but you start by saying shouldn't do, which sounds like you were disagreeing about something.

If i float in space with my bike and front wheel is spinning, i turn front wheel to the left, bike and i start rotating (leaning) to right. turning front wheel feels heavier because part of the force goes in rotating me and the bike.

Edit: in car all (most) of the steering force goes only to rotating the wheels around vertical axle so the wheels can be turned with less force, because part of it doesn't go to something else.

 

I think that what it is actually going on is that when steering torque tries to alter camber it is also affecting the eccentricity of tire load provoked torque which is the one actually providing FFB.

Hence this is a complex system which cannot be studied independently for both axes. Considering that full tyre dynamics are being considered in this sim I have no doubt now that the real effects causing FFB are properly implemented.

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Atleast it seems like the ffb is not as far from the truth as i thought before.

 

ok so that was said in context of the off-ground scenario (in which i used the extreme example of in being in space to also take gravity out of the equation but it's all the same). So no matter how fast the front wheel is spinning, turning/steering the front wheel will not result in a torque (though technically an angular momentum change and not a torque) that can be felt through the steering axis by the rider. The simplest reason being the resultant action from the steering input occurs 90 degrees later/out of phase (due to gyroscopic effect) which is not along the steering axis of rotation but perpendicular to it, hence you can't/won't feel it.

Shouldn't do. As i mentioned above, the resultant torque (or technically angular momentum change) occurs 90 degrees out of phase with the steering axis so you should not be able to feel it.

But when the tyre is on the ground and gravity is present, it can be "felt" due to how the gyroscopic effect/action causes a rotation of the tyre (i.e. in the camber direction) and in combination with things like caster angle of the steering tyre, it causes the production of a more restoring action from the caster trail. This is the same as why bikes can stay up on their own, remain stable and straight with sufficient speed. The gyroscopic effect/action is partly why they become more stable with increasing speed but it's not without the restoring effect from caster trail. The combination is what causes the steering wheel to feel heavier.

Check out this cambridge university piece on "...the insignificance of the gyroscopic effect when riding a bicycle
": http://www2.eng.cam.ac.uk/~hemh/gyrobike.htm

 

But, If you turn a spinning wheel (of the ground) and it also leans because of gyroscopic effect, part of the force (energy) you put in turning the wheel goes in leaning the wheel. So to get a wheel that is not spinning turned some amount takes some amount of energy. Then getting a spinning wheel turned the same amount takes a bit more energy, because part of it goes in to leaning the wheel.

 

hmm, that's an excellent point you've made. You must be right and i'm totally wrong. One can imagine that as the rotating speed tends towards infinity, no amount of input steering torque will result in the steering angle changing. But instead all of the input energy being channelled 90 degrees out of phase.

 

The question is whether the forces being discussed are significant enough to worry about in the scheme of things.

http://www.todayifoundout.com/index...thing-to-do-with-your-ability-to-ride-a-bike/

Personally I'd suspect that the resistance you feel due to gyroscopic forces would be completely dwarfed by everything else unless you isolated those forces in a specific, and unrealistic (for normal driving) scenario. But I'm certainly no expert

 

I'm not an expert neither but i think that the conclusion now is that gyroscopic effect doesn't have much influence on car's steering because car's front wheels doesn't have much of leaning movement, so pretty much all of energy put in to steering goes in turning the front wheels and not into leaning movement of the wheels.

In bicycle gyroscopic effect might not come in to play because of slow speeds (wheels rotates slowly), But i think that in motorcycles it starts to have effect making turning front wheel heavier and causing a force that helps motorcycle to lean. And there is still also the forces that has effect when riding the bicycle.

 

The fact that the torque appears 90 degrees to the leaned angle makes it consume no energy since the projection of the torque vector to the angular rotation vector is zero.

It basically is the same case as when force is perpendicular to displacement.

Weight when sliding a mass over a horizontal surface absorbs or provides zero energy since displacement is normal to the force. Friction force would absorb energy since it is in the same line but opposite sense to displacement.

Another case would be the centripetal force in a circular movement of a mass which actually changes direction of movement but is providing no energy since it acts perpendicular to the speed vector and hence maintaining speed magnitude.

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Non spinning/rotating wheel - all the steering input energy (neglecting energy losses from the system) results in direct transfer to steering axis rotational kinetic energy.

But take a spinning/rotating wheel - some amount of the input energy will cause an increase of rotational kinetic energy about the steering axis (just like the non spinning wheel scenario) but some will also be diverted to a rotation axis perpendicular to the steering axis due to gyroscopic effect.

In the hypothetical extreme case of infinite wheel spin, non of the steering input energy should result in an increase in rotational kinetic energy about the steering axis but instead all into a perpendicular axis about the steering axis. I.e. it should be purely gyroscopic effect/action only.

I'm curious how the energy is divided into the two with changing wheel speed and if difference in the power of the input (i.e. different rates of energy for the same total amount of energy input) affects the distribution of the energy input into the steering axis and one perpendicular to the steering axis.

 

I disagree with your last statement. You are considering that since a gyroscopic torque appears, it is absorbing energy. For sure the steering system has its own inertia that needs to be overcome to turn the steering wheel. That inertia opposes input steering wheel rotation and absorbs energy. However the gyroscopic torque does not absorb any any energy since as I said is perpendicular to the input angular displacement.

Therefore all the input energy is absorbed by the FFB resisting force which is the sum of the steering system inertia plus the effect of tire loads on the steering column.

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In car gyroscopic effect doesn't absorb energy because leaning movement can't happen. Not much atleast. If wheel spins with infinite speed, and it would be mounted infinitely rigidly to something where it can be rotated around one axle, like cars front wheel, turning the wheel would take same effort as turning wheel that is stopped. If wheel is not infinitely rigidly mounted happens what DrR1pper sayed. Thats my opinion at the moment atleast.