ever heard the "you can't race a honda up a hill" myth?
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ever heard the "you can't race a honda up a hill" myth?
I don't know if you've ever heard this myth or not, but sometimes when talking about hondas and their "low torque" people will mention you can't race a honda up a hill cuz they'd be so slow cuz of their "low torque".
Let me use an example. According to their logic, if you had a civic race a focus up a hill, the focus would easily win cuz it makes more torque, which you need to go up a hill.
Actually though, this is completely incorrect. The advantage in this case actually goes to the CIVIC because its lighter! Going up a hill does NOT change the amount of power you are putting out. The amount of force power you are making to move the car forward does not change.
However, since the focus is heavier, it will actually be slowed down even MORE than the civic. And the race advantage up the hill goes to the "low torque" civic.
So next time some a-hole tells you how slow your honda would be racing up a hill, you can tell them that racing up a hill is gonna slow them down even more than you if they have a heavier car.
Let me use an example. According to their logic, if you had a civic race a focus up a hill, the focus would easily win cuz it makes more torque, which you need to go up a hill.
Actually though, this is completely incorrect. The advantage in this case actually goes to the CIVIC because its lighter! Going up a hill does NOT change the amount of power you are putting out. The amount of force power you are making to move the car forward does not change.
However, since the focus is heavier, it will actually be slowed down even MORE than the civic. And the race advantage up the hill goes to the "low torque" civic.
So next time some a-hole tells you how slow your honda would be racing up a hill, you can tell them that racing up a hill is gonna slow them down even more than you if they have a heavier car.
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Rep Power: 0 I dont see the logic in it.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
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Originally Posted by NoFriends
I dont see the logic in it.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
i understand what s2kman is saying but im confused
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Rep Power: 400 the reason being, is because acceleration is directly affected by mass. once you go uphill, gravity multiplies the effect that mass has on your acceleration. this is because the force your car outputs now must use some of it's power to also fight the effect gravity has on the mass of your car. you use this formula:
mass*gravity*height. this determines the gravitational potential energy of the car.
I found this on a thread on s2ki:
m = mass (units Kg)
a = acceleration (units m/s^2)
f = dissipative forces (aerodynamic, loss in transmission, etc..) (units Newtons)
theta = incline angle, if you are going up a hill
g = accelration due to gravity (units m/s^2)
F = Net force that the ground exerts on the wheels (units Newtons)
then
m a = F - f - mg sin(theta)
Acceleration rate changes if you are going up an incline since the gravitational force mg sin(theta) depends on mass.
obviously the more mass you have, the greater effect is has on your car accelerating up an incline, since it increases the gravitational force.
mass*gravity*height. this determines the gravitational potential energy of the car.
I found this on a thread on s2ki:
m = mass (units Kg)
a = acceleration (units m/s^2)
f = dissipative forces (aerodynamic, loss in transmission, etc..) (units Newtons)
theta = incline angle, if you are going up a hill
g = accelration due to gravity (units m/s^2)
F = Net force that the ground exerts on the wheels (units Newtons)
then
m a = F - f - mg sin(theta)
Acceleration rate changes if you are going up an incline since the gravitational force mg sin(theta) depends on mass.
obviously the more mass you have, the greater effect is has on your car accelerating up an incline, since it increases the gravitational force.
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Rep Power: 0 I can see where you are coming from, but its still not clicking in my head.
I would love to test this out with a V6 camaro or something that is fairly even on flat ground but with more torque and a greater mass.
I would love to test this out with a V6 camaro or something that is fairly even on flat ground but with more torque and a greater mass.
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Originally Posted by NoFriends
I dont see the logic in it.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
We proved on the moon with a hammer and a feather that objects of different mass fall at the exact same rate with the absense of an atmosphere to affect drag and terminal velocity.
Now that being said, gravity can not be pulling "harder" on an object of greater mass, its only the aerodynamics of that can affect its fall rate. So while nothing of the cars have changed, only the incline of their racing surface, Im failing to see how one car would have the greater advantage.
They both have the exact same power to weight ratios that they have on flat level ground, and both are subject to the same new incline. They are both suffering the same penalties from going against gravity, and as their power to weight ratio hasnt changed, I dont see how the lighter car has the advantage.
ohohoh...no, no, no
when a car is on a totally flat surface, the gravity force vector (-y direction), g=m*a will cancel out with the normal force pointing at (+y). Therefore the movement of the car is purely depend on the kinetic friction of the tire, that is, directly associate to the torque t=d*f.
When the car is on an inclined surface, the gravity force vector will have 2 component, (-y and -x) the -x will be gx=-m*a cos(angle) where angle is the incline surface versus flat surface. A greater mass will definitely result a greater -x component therefore a smaller over all +x force.
please remember all what newton was saying is that 2 objects of different mass will have the same acceleration on a given planet, therefore "land" at the same time; however, the force excercice on them is totally governed by the 3rd law f=m*a...don't confuse the two.
A free body diagram would help us understand well here. Please correct me if I am wrong, but I believe my degree won't fail me here.
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Originally Posted by orion_squall
ohohoh...no, no, no
when a car is on a totally flat surface, the gravity force vector (-y direction), g=m*a will cancel out with the normal force pointing at (+y). Therefore the movement of the car is purely depend on the kinetic friction of the tire, that is, directly associate to the torque t=d*f.
When the car is on an inclined surface, the gravity force vector will have 2 component, (-y and -x) the -x will be gx=-m*a cos(angle) where angle is the incline surface versus flat surface. A greater mass will definitely result a greater -x component therefore a smaller over all +x force.
please remember all what newton was saying is that 2 objects of different mass will have the same acceleration on a given planet, therefore "land" at the same time; however, the force excercice on them is totally governed by the 3rd law f=m*a...don't confuse the two.
A free body diagram would help us understand well here. Please correct me if I am wrong, but I believe my degree won't fail me here.
when a car is on a totally flat surface, the gravity force vector (-y direction), g=m*a will cancel out with the normal force pointing at (+y). Therefore the movement of the car is purely depend on the kinetic friction of the tire, that is, directly associate to the torque t=d*f.
When the car is on an inclined surface, the gravity force vector will have 2 component, (-y and -x) the -x will be gx=-m*a cos(angle) where angle is the incline surface versus flat surface. A greater mass will definitely result a greater -x component therefore a smaller over all +x force.
please remember all what newton was saying is that 2 objects of different mass will have the same acceleration on a given planet, therefore "land" at the same time; however, the force excercice on them is totally governed by the 3rd law f=m*a...don't confuse the two.
A free body diagram would help us understand well here. Please correct me if I am wrong, but I believe my degree won't fail me here.
degrees are overrated.
j/k 
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ideal experiment would be 2 identical civic, one totally loaded with "mass" and the other totally empty.
Both drive up at the same torque (same gear, same force apply to petal etc etc)...you will see the empty one goes up faster...actaully it is so obvious that we don't even need to do such experiment, haha.
ideal experiment would be 2 identical civic, one totally loaded with "mass" and the other totally empty.
Both drive up at the same torque (same gear, same force apply to petal etc etc)...you will see the empty one goes up faster...actaully it is so obvious that we don't even need to do such experiment, haha.
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Originally Posted by Civic_RedLine
so how about racing a rwd car with our civic. Doesnt fwd pull harder and the rear loses power?
i go up a hill everyday when i come home cuz i live at the top of a hill and i never have a problem going up it. if i keep my foot on the gas at a constant speed of say 50 the hill slows my speed down to 40 just because of the incline. doesn't feel sluggish if i wanna push it more.
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Originally Posted by Civic_RedLine
so how about racing a rwd car with our civic. Doesnt fwd pull harder and the rear loses power?
as already stated, the amount of force power being put to the wheels on either car does not change just cuz you're going up the hill.
it's the amount of energy that must be diverted from that power to move the mass of the vehicle that is changing.
