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prometheum
 
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unassisted, discounting resistance, and all other things being equal besides weight, both cars will accelerate downhill at the same rate would they not?

i don't see how uphill, downhill, flat it makes a difference

edit: i think you might have an easier time beating a turbocharged car downhill though since he definitely won't be making full boost in first or second gear, and depending on how steep the hill is third
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Last edited by prometheum; 01-15-2007 at 05:59 AM..
Old 01-15-2007, 05:51 AM prometheum is offline  
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#31  

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prometheum
 
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Quote:
Originally Posted by Fear Night View Post
But the force that pulls the car is multiplied by mass. This is why it's harder for a heavier car to go uphill because there is a greater force working against it.

F=ma or (Force of gravity) = (mass of car)*(acceleration of gravity)

For downhill yes they both get the same benefit. Two similar shaped objects when dropped will reach the ground at the same time, regardless of mass. The only big thing taken into account is air resistance (which I highly doubt Rang3 had the advantage on vs a Mazda 3 ).

it doesn't matter, it exerts more force because it weighs more

drop a feather and a coin in a vacuum, they fall at the same rate because the force is directly proportional to the added weight.
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Old 01-15-2007, 05:54 AM prometheum is offline  
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#32  

mike27
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@ people who think that increased mass = increased acceleration from gravity
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Old 01-15-2007, 05:58 AM mike27 is offline  
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sun_ofa_beach
 
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wow the stupidity is strong in this thread
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Old 01-15-2007, 05:59 AM sun_ofa_beach is offline  
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punisher
 
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Originally Posted by jealousKEY View Post
@ people who think that increased mass = increased acceleration from gravity

Yup. Force of gravity on an object is proportional to mass, inerta is proportional to mass as well. I do hope most of the "retard" posts in this thread are due to ignorance rather than stupidity. Ignorance is fixable, stupidity is not.
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Old 01-15-2007, 06:49 AM punisher is offline  
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#35  

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at all the people that don't know kingergarten physics
Old 01-15-2007, 08:53 AM bob101 is offline  
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#36  

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Old 01-15-2007, 09:17 AM VanFanel is offline  
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#37  

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at all the people that don't know kingergarten physics


indeed, yes. I myself did learn physics in kindegarten- infact I was #1 in my class.


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Old 01-15-2007, 09:34 AM MrJohnson is offline  
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#38  

colinshark
 
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Your weight won't help you to any measurable degree.


Also, your "torque" only determined whether you could climb the hill or not. As soon as you started talking about "pulling on him" over a period of time, it's horsepower that won it for you. He probably didn't shift and fell out of his power band.
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Last edited by colinshark; 01-15-2007 at 10:36 AM..
Old 01-15-2007, 10:29 AM colinshark is offline  
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#39  

JimmySal
 
F=/=ma

F=m*dv/dt

noobs.
Old 01-15-2007, 10:49 AM JimmySal is offline  
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#40  

g
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To the OP, what car do you drive? Just out of interest I don't actually know, but I know you like older cars
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Old 01-15-2007, 11:40 AM    g is offline  
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Quote:
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To the OP, what car do you drive? Just out of interest I don't actually know, but I know you like older cars

'71 Pontiac Bonneville w/455(7.5L).. about 300hp/400tq as equipped now(originally a 2bbl w/ 280hp/380tq, now with 4 bbl ), but 4500++ pounds to push. It's not fast by any means, and on a level, I'm confident that the Mazda would have beaten me. That's why I posted this little question, because I pulled so easily ahead of him. Someone suggested he wasn't in his power band, but I disagree, he didn't shift any too early into second, and it sure sounded like he was going for it. He obviously doesn't have the low end torque, but he has much less weight, closer gear ratios, surely a more advantageous differential ratio and a transmission that doesn't soak up 40% of the power like mine probably does. In most circumstances like this, I would have pulled slightly ahead right off the line, then I would fall steadily further back once they got up in their power band. Meh, guess those Mazdas really are slow as shit
Old 01-15-2007, 11:53 AM Rang3find3r is offline  
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#42  

Fear Night
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Quote:
Originally Posted by JimmySal View Post
F=/=ma

F=m*dv/dt

noobs.
The derivative of velocity IS acceleration

dv/dt = a

Therefore, F=ma.
Old 01-15-2007, 11:55 AM Fear Night is offline  
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#43  

g
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300hp/400 lb-ft over a two tonne car isn't too disasterous, it'll certainly pull you up hills though


Cool car... I like those cars as big lazy cruisers
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Old 01-15-2007, 11:56 AM    g is offline  
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#44  

lopoetve
 
Quote:
Originally Posted by Fear Night View Post
But the force that pulls the car is multiplied by mass. This is why it's harder for a heavier car to go uphill because there is a greater force working against it.

F=ma or (Force of gravity) = (mass of car)*(acceleration of gravity)

For downhill yes they both get the same benefit. Two similar shaped objects when dropped will reach the ground at the same time, regardless of mass. The only big thing taken into account is air resistance (which I highly doubt Rang3 had the advantage on vs a Mazda 3 ).
Correct. but we're talking about downhill.
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Old 01-15-2007, 11:58 AM lopoetve is offline  
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