Originally posted by slade
uhh... what?
you didnt really even say anything in that post. you verbally restated F=ma as a=F/m.
then you use a diagram to talk about lift forces on the tires. okay, sure, as you defined it, a "lift force" is a normal force between the ground and the tire, which is specific to this situation. the terminology really doesnt matter. and your comparison is weaker than a honda nighthawk. the "lift force" you described in this situation has nothing to do with an airplane. An airplane doesn't fall out of the air because the coanda effect produces lift as the air flows over the curved wing. the car doesnt fall through the ground because... well, the wheels are on the ground.
and then you used the term weight transfer again, without even defining it. the term implies that weight is being transfered, which would effect the acceleration needed to lift the front wheel (for example, if the rider leaned forward or backward). what you were referring to is the redistribution of weight on the tires due to acceleration.
This is just another way of implying the exact same concept. Only in the automotive/racing field, "weight transfer" is the commonly used term. I'm having a hard time figuring out why don't feel the term "weight transfer" is appropriate. We've established the fact that we're referring to the same principle.
which is... exactly what I explained, without using the term "weight transfer". the bike accelerates, applying a force at the rear wheel. this force is behind and below the center of mass of the bike. that force will exert a torque on the bike, and if the acceleration is great enough, the front wheel will lift. if you care to rephrase that, you could easily say that a torque applied at the rear wheel will cause the distribution of force due to gravity on the vehicle will shift from the front wheel to the rear wheel, and if the force shifts completely the front wheel will lift. my explanation explains yours. if you cared do, and had all the correct values, you could set up a simple physics equation to calculate the force required on a certain bike to lift the front wheel.
I obviously didn't and it really is irrelevant. We were debating the term "weight transfer" and how it does/doesn't apply to this situation. We weren't trying to
calculate the force required to lift the front wheel of a bike.
my explanation is fundamentally the same as yours, except its not grounded in one scenario, and i didnt just copy it from a website. my explanation accounts for why a rear wheel drive vehicle can do a wheelie. it also accounts for why a front wheel drive vehicle cannot do a wheelie, and how the front of a hard-breaking vehicle will "dive", or how a vehicle that stops very hard can do an endo, and why, theoretically, if you strapped a rocket to the back of your car above its center of mass, your car could not do a wheelie. but you apparently didnt understand my post enough to get that.
Actually that's not the case at all. You're not denying anything that I have stated, only refuting my word selection and the term I used to describe the exact same concept. Whether you want to call it "redistribution of weight" or "weight transfer," it really desn't matter. We're not talking about adding or taking away physical weight.
you didnt really even say anything in that post. you verbally restated F=ma as a=F/m.
then you use a diagram to talk about lift forces on the tires. okay, sure, as you defined it, a "lift force" is a normal force between the ground and the tire, which is specific to this situation. the terminology really doesnt matter. and your comparison is weaker than a honda nighthawk. the "lift force" you described in this situation has nothing to do with an airplane. An airplane doesn't fall out of the air because the coanda effect produces lift as the air flows over the curved wing. the car doesnt fall through the ground because... well, the wheels are on the ground.
and then you used the term weight transfer again, without even defining it. the term implies that weight is being transfered, which would effect the acceleration needed to lift the front wheel (for example, if the rider leaned forward or backward). what you were referring to is the redistribution of weight on the tires due to acceleration.
This is just another way of implying the exact same concept. Only in the automotive/racing field, "weight transfer" is the commonly used term. I'm having a hard time figuring out why don't feel the term "weight transfer" is appropriate. We've established the fact that we're referring to the same principle.
which is... exactly what I explained, without using the term "weight transfer". the bike accelerates, applying a force at the rear wheel. this force is behind and below the center of mass of the bike. that force will exert a torque on the bike, and if the acceleration is great enough, the front wheel will lift. if you care to rephrase that, you could easily say that a torque applied at the rear wheel will cause the distribution of force due to gravity on the vehicle will shift from the front wheel to the rear wheel, and if the force shifts completely the front wheel will lift. my explanation explains yours. if you cared do, and had all the correct values, you could set up a simple physics equation to calculate the force required on a certain bike to lift the front wheel.
I obviously didn't and it really is irrelevant. We were debating the term "weight transfer" and how it does/doesn't apply to this situation. We weren't trying to
calculate the force required to lift the front wheel of a bike.
my explanation is fundamentally the same as yours, except its not grounded in one scenario, and i didnt just copy it from a website. my explanation accounts for why a rear wheel drive vehicle can do a wheelie. it also accounts for why a front wheel drive vehicle cannot do a wheelie, and how the front of a hard-breaking vehicle will "dive", or how a vehicle that stops very hard can do an endo, and why, theoretically, if you strapped a rocket to the back of your car above its center of mass, your car could not do a wheelie. but you apparently didnt understand my post enough to get that.
Actually that's not the case at all. You're not denying anything that I have stated, only refuting my word selection and the term I used to describe the exact same concept. Whether you want to call it "redistribution of weight" or "weight transfer," it really desn't matter. We're not talking about adding or taking away physical weight.
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