but would convert to pounds
force (675N is about 151 lb).
remind me not to let anyone here do my math for my gearing, you guys throw
these long math problems up here all technical and then use a statement of
"about", well your "about" figure is off by 30.87 lb Simple math errors can
have real shitty end results maybe your pencils are dull :)
"nothing like racing on paper" its kinda like ordering all the bolt on
horsepower parts out of a catalog
just having fun with the post guys , not trying to slam anyone
dan
From: Mark Osterbrink <[EMAIL PROTECTED]>
Reply-To: [email protected]
To: [email protected]
Subject: Re: [F500] Torque at 150 MPH
Date: Wed, 02 Aug 2006 20:06:24 -0400
You were supposed to do it with LESS than 72 hp :)
In short, at any speed you are traveling over a distance in a time. While
you travel that distance, you are applying a force to overcome drag. Force
applied over a distance is work, and work per time is power. Power
determines speed, maximum power determines top speed.
You have the drag equation right, but have a few errors from there.
1/2*CD*A*rho*V^2 = 0.5*0.6*0.5*1.2*67.056^2 = 809 N of drag force.
Newtons don't convert to foot-pounds, but would convert to pounds force
(675N is about 151 lb).
809 Newtons at 67 m/s = 54273 Watts, or 72.78 hp
Can't go that fast with that car with less than 72 hp.
Your example (72 hp @ 2500 RPM, 1.0 gearing and 10.1" tire radius) gets you
there within rounding errors, but that's because you started with the 72 hp
limit I gave you.
If we take your 151 ft*lbf assumption, and apply it to 50 hp instead of 72
you can see the problem. 151 ft*lb at 1739 RPM is 50 hp. 151 ft*lb times
the 1.0 gear ratio divided by the 10.1" radius gives us 809 N of force, so
we have enough force. But 1739 RPM times 1.0 gear ratio times 10.1" times
2pi only give us 104 MPH. If we change the gearing to give us 150 MPH, we
only have 556 N - not enough grunt. Can't get there with only 50 HP.
Like I said, power is what's important, torque was invented to confuse and
misdirect :)
Chuck Voboril wrote:
Red Doggie,
Don't really need a steam engine afterall.
Assuming zero acceleration at steady state speed (and no mistakes in the
SI vs English conversions)
Using classic equation for drag force in Newtons using your Cd, frontal
area, density of the medium, and speed in m/s.
N=1/2*Cd*A*rho*v^2
I came up with 675.4 Newtons SI which converts to 151 ft-lbs. English
Then, a motor which produces 72 HP at 2500 RPM will have the required 151
ft-lbs.
10.1" radius tires, 1:1 gearing at 2500 rpm
speed(MPH)=RPM*tire radius/(168*gearing)
You do have a very valid point that HP cannot be ignored.
It is totally tied to HP and RPM.
However, since RPM was free, I used it to my advantage.
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