Centrifugal forces

[faffi]

Since we've had the recent rejuvenated debate of spring rates, I want to bring to attention another topic where what at least I feel - like with suspension preload - and what physical laws say contradict.

45 degrees of lean angle should equal 1G of cornering force. However, in my experience, it is possible to reach deeper lean angles at low speeds than at high speeds. As an example, MOTORRAD managed to lean a BMW K75 51 degrees at a walking pace, but the bike does not have enough cornering clearance to master that at higher speeds.

So here is the questions: Will the tires offer more grip for any degree of lean at 10 mph than at 100 mph? Will the suspension compress less at 10 mph than at 100 mph with the same amount of lean?

[Huzo]

A 45 degree turn has 1.4 g of force down through your spine, it is the square root of 2.
Pythagoras was talking to me about it yesterday….
If you draw a diagram with a vertical line say, 1 foot and a horizontal line at the bottom at 90 degrees of (also) one foot, you will have something that looks like a half a square.
The vertical line can represent your weight of (say) 400 kg total which is gravitational and pointing towards the Earth’s core.
The horizontal line is the centripetal force vector of the same magnitude.
A diagonal line joining the top of the vertical and end of the horizontal lines will be 1.4 x 400 = 560 kg.
If you weighed 100 kg when you jumped on your bike, you will “feel” 141 kg through your ass in that 45 degree bank. (Lean).
Same in a plane during a 45 degree bank.
BTW.. In the turning diagram, he is looking the wrong way, but I digress.

To directly reference your question.
When our hero leans his bike to 45 degrees, he will be holding his C of M “outside” of the turn, such that the C of M of the entire combination, is somewhere between his body and the bike.


Also.
Force trying to skid the tyre is also a function of velocity.
Also the suspension will compress more at 100 mph than 10, it is the road surface providing the equal and opposite force.
In conclusion…
As a precursor to what will doubtlessly become a raging debate, including such incalculably valuable anecdotes as…
“My buddy used to have a Norton…….” etc….
It might be worth reading up on the definition of Centripetal and Centrifugal, that will iron out a few creases.

[faffi]

I have tried, and it makes my brain hurt every time

Anyway, 1G is what is the thing I consider to be cornering force (even if label it wrong), the 1,41G going through my spine and arse is - for me - of less interest, because it is the speed that matter

Also, if I understand this correctly, this theoretical cornering force is depending on tires of zero width. The wider the tire, the less speed for any given speed. Also, since tires have width, CoG also matter; the lower the center of gravity, the more you must lean for any given speed. Meaning that even if a low-sitting cruiser with wide tires could be leaned over 45 degrees, it would not manage 1G of cornering force, is that correct? And that with zero width tires, the CoG would be irrelevant, meaning you would need to lean the same amount, regardless of where the CoG is placed?

[Tom H]

Huzo I believe your wrong on how much weight your arse would feel!

I believe that you would feel, in an upright position, 100KG minus the weight of your legs and some of the weight of your arms then use the multiplier to come up with the correct number.

I do feel that the more you lean forward to the bars, the weight on you rear would decrease and increase on your arms. But the more forward you lean and due to the wind against your body, the wind might actually take weight off your arms and rear due to lift from your body.

All I gots to say,
Have fun!
Tom

[Huzo]

Huzo I believe your wrong on how much weight your arse would feel!

I believe that you would feel, in an upright position, 100KG minus the weight of your legs and some of the weight of your arms then use the multiplier to come up with the correct number.

I do feel that the more you lean forward to the bars, the weight on you rear would decrease and increase on your arms. But the more forward you lean and due to the wind against your body, the wind might actually take weight off your arms and rear due to lift from your body.

All I gots to say,
Have fun!
Tom

I do agree that if you don’t mind a bit of mud in the water, we can try to allow for all that.
I think we would both agree that Faffi is learning to walk here, so I was hoping not to drag him into learning the intricacies of a Russian ballet.
But to address your point…
How about if I say that the TOTAL weight bearing parts of your body will experience an increase of 41.4% in the load they bear.
Also Faffi, the width of the tyre has no bearing, it is the contact patch that is experiencing the force vectors and the contact patch must ALWAYS be directly under the g forces eminating from the centre of mass.
Whether the bike is upright or leaning, the resultant of all vectors will be a straight line from the c of m through the contact patch.
If it were not thus, there would be a resultant horizontal vector acting and you would immediately crash your guts out…..

[faffi]

Hm. But the wider the tire, the further the contact patch will move away from the center line of the bike, creating a vector(?) that must be compensated for by more lean for any given speed, no?

Put it another way: if I want to take a corner at 50 mph, I must lean my bike more the wider the tires are. Also, the lower the CoG is, the more I need to lean the bike. I can compensate some, or perhaps all of this by moving the CoG up and/or to the inside of the corner. At least this is what I have picked up, even if I cannot give you the mathematics for it.

[Huzo]

Hm. But the wider the tire, the further the contact patch will move away from the center line of the bike, creating a vector(?) that must be compensated for by more lean for any given speed, no?

Put it another way: if I want to take a corner at 50 mph, I must lean my bike more the wider the tires are. Also, the lower the CoG is, the more I need to lean the bike. I can compensate some, or perhaps all of this by moving the CoG up and/or to the inside of the corner. At least this is what I have picked up, even if I cannot give you the mathematics for it.

Hey Faff..
I do concede to your point regarding tyre width and its lateral distance away from the centreline. I should have allowed for that.
The point regarding the height above the point of support (ground), of the c of m is worth drilling into.
The height of the c of m has no bearing on lean angle, as long as the c of m is a straight line to the contact patch, the force trying to throw the bike to the outside of the turn, is equal to the force trying to throw it on it’s side.


You can perhaps appreciate, that if the c of m is way up high, then there is a lot of leverage trying to throw you “outside” but also that comes with a lot of leverage trying to drag you down.
They always balance regardless of how far up that 45 degree line you go.
Remember a vector is defined by its sense of direction and the amount of force is determined by its length.

[aklawok]

 Cornering rate vs/lean angle is a basic and instinctive equation. Mathematical interpretation will only befuddle. Enter low...exit high...that is it!!

[Huzo]

Cornering rate vs/lean angle is a basic and instinctive equation. Mathematical interpretation will only befuddle. Enter low...exit high...that is it!!

So can you enlighten us who are so easily befuddleable, what that instinctive equation might be ?
You know, I know, faffi knows and everyone including Donald Duck, knows that a bike leans into a corner, but can YOU tell me why ?
I mean….WHY ?
Also do you think for one moment that you get to determine how much lean angle you will use ?
Now something I do need educating on, what does “Enter low…exit high” mean ?

[Vagrant]

Faffi, it's time to change your name to "Instigator"!

[faffi]

Anything to please you, Vagrant.

Here is an article I found, unfortunately in German, but google translate + the pictures should make it accessible for everybody who find the topic interesting. It does show and explain the influence of tire width and center of mass.

PS! Click on the left option to proceed if that window opens, the right is for paid access.

https://www.motorradonline.de/ratgeber/alles-ueber-schraeglage-schraeg-schraeger-am-schraegsten/

[73 sport]

   You can also use a Centipedal Arc Equation for gross results. To get a more precise answer, use a Millipedal Arc Equation. The above calculations are invalid if rider is prone or supine. Above all , don't face the wrong way in a turn.

[SIR REAL ED]

   You can also use a Centipedal Arc Equation for gross results. To get a more precise answer, use a Millipedal Arc Equation. The above calculations are invalid if rider is prone or supine. Above all , don't face the wrong way in a turn.

Why would any self-respecting MotoGusti care about the corning speeds, lean angles, or G-forces experienced by a Centipede or a Millipede?   

Lets keep this thread focused on motorcycling or some well intentioned wanker will complain to a moderator.....   

[SIR REAL ED]

https://sciencenotes.org/centrifugal-force-definition-formula-examples/

[aklawok]

Give engineers a ton of money to solve the problem: can we teach a robot to ride a motorcycle? Yes, and Yamaha did it, at race speeds.
https://youtu.be/mafJmMGGOXk?si=_kmeyshcAvG-0uYB
By eliminating human variables, Yamaha wanted real world repeatable data on performance and handling.
But at that time in 2017, the instinctive nature of Carlos rossi still beats it. All the math in the world and the computer still needs to "learn" by trial and error....just like a human.

[chuck peterson]

Words, words, words….

A picture is worth a thousand…for instance..

[DenverSteve]

   You can also use a Centipedal Arc Equation for gross results. To get a more precise answer, use a Millipedal Arc Equation. The above calculations are invalid if rider is prone or supine. Above all , don't face the wrong way in a turn.

Maybe things are different in Maryland but I've never entered a turn riding supine or prone.  Exiting a poorly calculated turn at high-speed..... is another story.