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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…. :rolleyes:
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 :violent1:
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 :evil:
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! :evil: :grin:
Tom
Huzo:
--- Quote from: Tom H on June 13, 2026, 06:03:09 PM ---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! :evil: :grin:
Tom
--- End quote ---
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…..
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