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Pit Stop:General Biking Discussion
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#51 (permalink) |
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Senior Member
Join Date: Mar 2007
Location: New Delhi
Posts: 272
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what to say. Sir OF is Oxford of biking and any technical knowledge.
bows down three times and then four times again INTO 50 times ![]() .
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#52 (permalink) |
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ElectroniX!
Join Date: Sep 2008
Location: BLR/GHY/MAS
Posts: 2,384
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Thanks for the great reply OF. I have some doubts. I will just frankly lay them out here.
Every action has an equal and opposite reaction. When we brake, a opposite torque is generated, which tends to press the front end downward and lift up the rear end. (Opposite to what happens during acceleration.) This force acts down on the front end. This force is related to the mass and moment of inertia of the bike. This causes a bike to dive. So far ok. But I feel this is not weight transfer. Weight transfer can occur only when there is a shift in center of gravity. A bike with no suspensions can not change its C of G by braking. Weight transfer, if allowed to occur by using a suspension, will allow a shift in C of G and thus contribute to further diving. And I thought loading up the front delivers more traction to the front wheel during a turn. So it is necessary. If not this, what exactly does trail braking achieve? Waiting to hear from you. Frankly it is a very interesting discussion.
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Your biking tells a lot about the person you are! Last edited by abhijeet080808; 06-28-2009 at 11:10 AM. |
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#53 (permalink) | ||||
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Senior Member
Join Date: Oct 2007
Location: Noida
Posts: 648
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c_sbk:
![]() A hundred thanks for the appreciation!!Quote:
All right…lets take this further… Taking your first statement: Quote:
When you brake, the bike tends to continue forward (that’s its liner inertia). But the front tyre resists forward movement (through friction) and since the point of application of the resistive force is not in line with the C of G of the rider/bike combo, a rotational force develops (its magnitude defined by rotational moment of inertia of the rider/bike combo). This rotational force tends to rotate the bike around the contact patch, tending to raise the C 0f G. Now going back, the linear force had a force component (F Cosθ) that acted downwards through the forks compressing them. The rotation of the C of G adds to this force. All this ‘results’ in weight transfer. The ‘weight transfer’ is a consequence of the force components. The origin is the force. It is not the other way round. The weight transfer does NOT generate the forces. It is the forces that cause the weight transfer. Your second statement: Quote:
Let us split the term ‘weight transfer’. Weight is nothing but a force (mg) where a certain mass is acted upon by the force of gravity (in literal sense that mass is being accelerated towards the centre of the earth at 9.8m/sec2) But since the ground below it is solid and able to resist this downward acceleration, an equal and opposite reaction manifests itself that we measure as ‘weight’. So ‘weight transfer’ is actually the path of the force acting through the front forks. It CAN move the C of G but this movement (raising or lowering) of the C of G is not the ONLY proof of its existence. The force is there whether or not the C of G moves. Of course the location of the C of G defines the magnitude of this force. The higher the CG, the greater is the rotational force. During hard braking, the CG gets raised due to fork dive allowing further ‘weight transfer’. This goes on till either 1.) The bike comes to a stop or 2.) The force exceeds the traction limit, the front tyre slides and the load gets removed. But in case your bike had anti-dive (a device that restricts fork dive/compression), you still get weight transfer to the front wheel. If it was not so, then anti-dive should actually reduce braking efficiency. Elaborating on the mechanism of braking further: Taking the example of a 70kg rider on a 150kg ZMA, the ‘weight transfer’ can be pretty easily approximated. It depends on the Braking Force and the ratio of the height of the CG to the wheelbase. Weight transfer = braking force x (C of G height/wheelbase) ratio Case 1 when C of G is assumed at 20 inches ZMA wheelbase: 1355mm = 1.355 m Height of C of G rider/bike combo = 20 inches = 508mm ~ 0.5m Ratio: 0.5/1.355 = 0.369 ZMA braking force (panic stop 1G deceleration) = 220 x 9.8 = 2156N Weight transfer = 1470 x 0.369 = 795N = 81kg So of the total braking force of 220kg (a 1G stop), some 82kg gets transferred to the front tyre. This is addition to the ‘static loading’ of the front tyre. What that means is that if we assume a 50-50 weight distribution over both tyres (though 50-50 is rarely true as the CG is rarely right in the middle), the front tyre is carrying a static load of 110kg. The 82 kg ‘additional’ weight transfer due to the forward motion of the bike takes the total load to 192kgs. Now 220 – 192 = 28kg..thats the load on the rear tyre. No wonder it slides so easily when you try to brake with the rear. Case 2: assume the CG higher by some 5 inches: New CG height: 25 inches = 635mm = 0.635m New ratio = 0.635/1.335 = 0.475 Weight transfer (case 2) = 2156 x 0.475 = 1025N = 104kg Total weight on front tyre: 104 = 110 = 214kg Weight on rear tyre = 220 – 214 = 6kg You are very very close to a stoppie!!! Just 6kg on the rear… And all this by just raising the CG by a mere 5 inches. No wonder the CG location is such a fight for motorcycle designers. Moral of the story: if you want to make a controlled panic stop without converting it into an inadvertent stoppie, tuck in lower on the bike when you brake hard. Now keep in mind the change in these numbers if the weight distribution is NOT 50-50 on both wheels but something like 45-55 as is usual in the real world. Abhijeet: I know this goes a trifle beyond what you asked but then understanding the physics of motorcycling is a lot more interesting when you get practical numbers to relate to. Quote:
Last edited by Old Fox; 06-30-2009 at 03:15 PM. |
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#54 (permalink) | |
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Senior Member
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Quote:
Just wonderful to see you spend so much time with figures, diagrams, formulae and examples, just for the benefit of understanding of another fellow biker. Coming back to the start of it all, I would like to highlight what OF said in the underlined statement. It is the loading of the front wheel that makes 'trail-braking' useful. IMHO, a softer suspension would * load the front wheel much lesser. * load up the front wheel with large amount of force suddenly I hope I didn't start it all over again. *Peace*
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#55 (permalink) | |
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Senior Member
Join Date: Oct 2007
Location: Noida
Posts: 648
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Quote:
. And I guess I have written enough on the subject to make for a small book. Suspension 'softness' or 'hardness' has no great relation to the weight transfer. Weight transfer is force. If the suspension is soft, it will compress it more, if it is hard, then the force can only compress it less. True, the suspension dive does alter the C of G of the bike but this alteration is not so substantial so as to have a huge impact on the force generated. This change in C of G does affect the handling of the bike. We have discussed this at length in my first response. Trail braking gives 2 major advantages: 1. Allows higher average speeds through the turn 2. Controlled loading of the front wheel to keep it within safe traction limits. Almost all of the questions that arise in your mind will have their answers in my three posts. I don't feel the need to actually point them out. look for them. they are there. As for anything new...I am game. |
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#56 (permalink) | |
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Senior Member
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Quote:
I have read through the posts two times. I think I am one time short. But again, there is a lot of automotive physics involved here, which is beautiful, frankly.
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#57 (permalink) |
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Senior Member
Join Date: Mar 2009
Location: Mumbai
Posts: 1,037
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What's "CG"?
Kidding . Couldn't resist that. Some awesome explanation there 'Old Fox', and 'abhijeet080808' too, for your perspective. It helps make things clearer.I too have a query. Was wondering how much would the angle of the fork have effects on the handling of the bike. Like some cruisers are angled quiet extremenly while normal street bikes are almost perpendicular. I unserstand, extreme angle would like cruisers would have longer wheelbase hence bigger turning radius, unflickable etc. But on a normal street bikes we used here, how much would the angle effect the handling characteristics? I'm also thinking, the contact patch of the tyre on the road will also differ due to angle. |
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#58 (permalink) | |
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Observer
Join Date: Oct 2008
Location: Bangalore
Posts: 59
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Quote:
You really explain really well! I have another question: What is the relationship between the weight of the bike with the width of the tyres? I used to think that the wider tyres are for heavier / more powerful bikes, but then what about the Bullets?
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Which of my bike rides is my favorite? The one I’m going to take tomorrow. My website: www.samudraNB.com |
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#59 (permalink) |
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Senior Member
Join Date: Apr 2009
Posts: 964
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awesome discussion going on
![]() unbelievable OF sir ...... noob sawal why do people crave for more cylinder engines ?? like suppose we have two bikes A and B of 250cc A with a single cylinder with same power and specs like B B with a twin cylinder n same specs like A then what does make a difference actually we are getting the same amount of power... same torque ,,, @ OF sir please could you elaborate this Trail Braking is a technique where the rider progressively reduces his braking force as he gets deeper into the turn and closer to the apex. i am not understanding this..... we keep the brakes pressed and slowly and steadily keep on releasing it is it what i am saying ??? then how will we maintain speed or for that matter even ride at a speed with more part of the brakes pressed or it is that i dont understand what u meant by BRAKING FORCE ![]() please explain sir
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#60 (permalink) |
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ElectroniX!
Join Date: Sep 2008
Location: BLR/GHY/MAS
Posts: 2,384
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Thanks for the awesome explanation OF. Cleared a lot of my doubts.
![]() One question - What are the factors that determines the ignition timings in a 4 stroke engine? How does the following factors affect timings - 1. Compression Ratio 2. Air-Fuel Ratio 3. Altitude 4. Temperature 5. Valve Timing And any other factors you may think off!!
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