HP needed for reaching 150MPH (model)
05-27-2001, 07:46 AM
#1
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HP needed for reaching 150MPH (model)
width 68.1in 1.72974 m
height 57.1in 1.45034 m
area (A) 2.508711112 m2
Drag Coefficient 0.35
Density of air(q) KG/m3 1.202 KG/m3
Velocity(kph) 150 241.35 km/hr
Wind Velocity 0
Power in kW p=(12.9x10**-6)*q*c*A*V*(V+WV)**2
p=(12.9x10**-6)*1.202*0.36*2.5087*241.35*(241.35+0)**2
Power on wheels required for tiby 191.41 kW 256.9203544 HP
150 MHP,sea level,no wind
flat terrain, considers no rooling
power,area calculated by height
times width (area is high due to
curvature which reduces cross
section area), estimated drag
coef for type of vehicle
[ May 27, 2001: Message edited by: Mario ]
height 57.1in 1.45034 m
area (A) 2.508711112 m2
Drag Coefficient 0.35
Density of air(q) KG/m3 1.202 KG/m3
Velocity(kph) 150 241.35 km/hr
Wind Velocity 0
Power in kW p=(12.9x10**-6)*q*c*A*V*(V+WV)**2
p=(12.9x10**-6)*1.202*0.36*2.5087*241.35*(241.35+0)**2
Power on wheels required for tiby 191.41 kW 256.9203544 HP
150 MHP,sea level,no wind
flat terrain, considers no rooling
power,area calculated by height
times width (area is high due to
curvature which reduces cross
section area), estimated drag
coef for type of vehicle
[ May 27, 2001: Message edited by: Mario ]
05-27-2001, 01:07 PM
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It has several assumptions which might change the numbers:
1)Area: I simply multiplied height by width but it should exact, the frontal largest cross sectional area.
2) Drag coef: I took the average of similar vehicles which I assume it has the rolling friction coef.. I couldnt find the exact number for the tiby. It could be lower.
But yes, you could assume the number is around that figure, give or take. You can plug the equation in exel and play with it if you consider the frontal area is less and the coef. is lower. There is a simple way to calculate your drag coef and rolling coef together so that this number is more realistic. The frontal area would need more work unless you can find it somewhere, I truly didnt looked for it too much but could not find it.
[ May 27, 2001: Message edited by: Mario ]
1)Area: I simply multiplied height by width but it should exact, the frontal largest cross sectional area.
2) Drag coef: I took the average of similar vehicles which I assume it has the rolling friction coef.. I couldnt find the exact number for the tiby. It could be lower.
But yes, you could assume the number is around that figure, give or take. You can plug the equation in exel and play with it if you consider the frontal area is less and the coef. is lower. There is a simple way to calculate your drag coef and rolling coef together so that this number is more realistic. The frontal area would need more work unless you can find it somewhere, I truly didnt looked for it too much but could not find it.
[ May 27, 2001: Message edited by: Mario ]
05-28-2001, 05:05 AM
#5
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QUOTE
Originally posted by Mario:
width 68.1in 1.72974 m
height 57.1in 1.45034 m
area (A) Â 2.508711112 m2
Drag Coefficient  0.35 Â
Density of air(q) KG/m3 Â 1.202 KG/m3
Velocity(kph) 150 241.35 km/hr
Wind Velocity  0 Â
 Â
Power in kW p=(12.9x10**-6)*q*c*A*V*(V+WV)**2 Â
p=(12.9x10**-6)*1.202*0.36*2.5087*241.35*(241.35+0)**2 Â
Â
Power on wheels required for tiby 191.41 kW Â Â Â 256.9203544 HP
150 MHP,sea level,no wind  Â
flat terrain, considers no rooling  Â
power,area calculated by height Â
times width (area is high due to Â
curvature which reduces cross Â
section area), estimated drag  Â
coef for type of vehicle
[ May 27, 2001: Message edited by: Mario ]
width 68.1in 1.72974 m
height 57.1in 1.45034 m
area (A) Â 2.508711112 m2
Drag Coefficient  0.35 Â
Density of air(q) KG/m3 Â 1.202 KG/m3
Velocity(kph) 150 241.35 km/hr
Wind Velocity  0 Â
 Â
Power in kW p=(12.9x10**-6)*q*c*A*V*(V+WV)**2 Â
p=(12.9x10**-6)*1.202*0.36*2.5087*241.35*(241.35+0)**2 Â
Â
Power on wheels required for tiby 191.41 kW Â Â Â 256.9203544 HP
150 MHP,sea level,no wind  Â
flat terrain, considers no rooling  Â
power,area calculated by height Â
times width (area is high due to Â
curvature which reduces cross Â
section area), estimated drag  Â
coef for type of vehicle
[ May 27, 2001: Message edited by: Mario ]
Listen to something.My tibby produces 249,8 at the wheels(turbo´d)and the top speed in my speedometer is 235-238(max)All this for your info.Note that 0-100kms in 6.7sec.Compare that with your results.C U.
05-28-2001, 05:38 AM
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All I have to say is pray your car doesn't have a limiter, my top is 210 kph ( 130 mph )and thats with a 55 shot of nos, and thats also at 5200 rpm. No limiter and I could easily hit 140 mph or so.
05-28-2001, 06:49 AM
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Well this is vehicle dynamics formulae which nobody can argue about. Only two variables could be wrong, the cross sectional area and the drag coefficient. Also altitude modifies the q for density of air. These variables can change the results. I again insist I could not find the drag coefficient so I took an average of similar cars from a mechanical engineering book and the area is less since I could not find the cross sectional area I just found the box area which is higher than reality. I do not intend to calculate it exactly, It is just to illustrate an aproximation to the power needed to reach such speed. You can find out your rolling and drag coefficient and find out the exact area and with the altitude where you race you can plug in the density of air and plug them into the formula that would be much closer to reality, or take a model car into a wind tunnel to find out ($$$$$$).
05-28-2001, 04:37 PM
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Time is indirectly involved in the velocity miles per hour. Look in the formula, if the testing road is not flat, or if you have a tail wind the HP drops, also higher altitude q air density willl decrease but it is also harder to produce the HP.
05-29-2001, 02:25 AM
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Well, there are a few misnomers...
The coefficient of drag on the old-school Tiburon is supposed to be 0.33, I don't know how they came to that number but that's what's posted.
The frontal area is harder to calculate, and taking the height times the width will give you the frontal area of a Tiburon Mack Truck. The front of the car isn't a complete square (height x width), it's the nose and the airdam (hood height x partial car width)...
I know there are cars that are bigger than the Tiburon with less than 250WHP that will make it to 150mph, and I think the only flaw in your calculation is your number for frontal area.
-Red-
The coefficient of drag on the old-school Tiburon is supposed to be 0.33, I don't know how they came to that number but that's what's posted.
The frontal area is harder to calculate, and taking the height times the width will give you the frontal area of a Tiburon Mack Truck. The front of the car isn't a complete square (height x width), it's the nose and the airdam (hood height x partial car width)...
I know there are cars that are bigger than the Tiburon with less than 250WHP that will make it to 150mph, and I think the only flaw in your calculation is your number for frontal area.
-Red-