what is your 8's top speed????
05-10-2007, 05:59 PM
#76
^obviously they dont require a highschool degree for an auto mechanics degree. You've done nothing but flame people ever since you came on to this site. Take it easy, not everything is a competition and we dont care how big your ***** are.
And for the record, there is no driver skill require on a top speed run.. youre saying you can do WOT better than I can? Wheres your logic?
4 people in the car and you claim 160? i claim BS unless you were on a downhill stretch.
p.s da stallion watch the personal attacks..
And for the record, there is no driver skill require on a top speed run.. youre saying you can do WOT better than I can? Wheres your logic?
4 people in the car and you claim 160? i claim BS unless you were on a downhill stretch.
p.s da stallion watch the personal attacks..
The only way it would EVER see 160 is if it flew off a cliff and was gravity assisted...altho I'm not sure what the terminal velocity of the car would be...
OR he was going down a long steep mountain road...and lunar gravity is on your side...and its approaching sea-level...and its like 50F outside...and your car magically found 50 more HP...
This is just another example ultimately of a desire for attention. Sort of how he got into the first pickle with the toe-gay fog drifter thing.
Last edited by eviltwinkie; 05-10-2007 at 06:02 PM.
05-10-2007, 06:01 PM
#77
Please don't beat me!
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I'm not going to make any personal jabs or try and take sides, but I would just like to see the video for my own amusement and "hope". P.S. Use sandbags to represent the people's weight to avoid a possible "bag full of sh*t on fire getting stomped out" kind of thing and only waste one life instead of four just in case.
05-10-2007, 06:08 PM
#78
05-10-2007, 06:10 PM
#79
05-10-2007, 06:46 PM
#82
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165 mph?! haha! 
05-10-2007, 07:06 PM
#83
A lot of this thread is bullshit.
1. Wind resistance on a stock 8 at that speed is something like 270 whp. That does not include rolling resistance. It could be done.
2. My A-Spec 8 is at 3500 rpms at 70 mph. That's 140 at 7k. Not sure how one could do 40+ more mph with 500 rpms.
1. Wind resistance on a stock 8 at that speed is something like 270 whp. That does not include rolling resistance. It could be done.
2. My A-Spec 8 is at 3500 rpms at 70 mph. That's 140 at 7k. Not sure how one could do 40+ more mph with 500 rpms.
05-11-2007, 09:39 AM
#85
05-11-2007, 10:09 AM
#88
http://www.rbracing-rsr.com/aerohpcalc.html
This isn't how I did it, but I found it interesting.
05-11-2007, 10:25 AM
#89
Drag Force = .5 * Cd * V² * A * unit vector *rho
Ignore the unit vector as we will assume linear motion, motionless fluid, and opposing drag force.
Power = Force * Speed
Drag Power = .5 * Cd * V³ * A * rho
I greatly prefer metric units for decent calculations.
Cd is dimensionless and I believe is .31 for an Rx-8
Rho is the density of the fluid. Air is generally listed as ~1.29 kg/m³ at 0 degrees Celsius. At common room temperatures it can be approximated by 1.20-1.23 kg/m³. This is due to the variation of density inversely with absolute temperature ratio.
V is (minus the unit vector) speed desired in meter/second. 1 mph is about .45 meters/second though I used a unit conversion program to do 185 mph to 82.7 m/s
A is the frontal area in m². 1 ft² is about .093 m². An rx-8 is about 52.8 inches tall and 69.7 inches wide. I assumed it took about 80% of that area up. That's 25.56 ft² or 1.90 m².
DragPower = .5*.31*82.7³*1.9*1.22
= 203157.2 watts
~ = 272.3 whp (1 hp, depending on definition, is about 746 watts)
Edit: Correcting for my left out variable, I again get 270 whp required which agrees with my initial post. In general, I try not to just pull **** out of my ***.
Edit: Took out comments about the incorrect calculation as this is now representative of how I did it.
Ignore the unit vector as we will assume linear motion, motionless fluid, and opposing drag force.
Power = Force * Speed
Drag Power = .5 * Cd * V³ * A * rho
I greatly prefer metric units for decent calculations.
Cd is dimensionless and I believe is .31 for an Rx-8
Rho is the density of the fluid. Air is generally listed as ~1.29 kg/m³ at 0 degrees Celsius. At common room temperatures it can be approximated by 1.20-1.23 kg/m³. This is due to the variation of density inversely with absolute temperature ratio.
V is (minus the unit vector) speed desired in meter/second. 1 mph is about .45 meters/second though I used a unit conversion program to do 185 mph to 82.7 m/s
A is the frontal area in m². 1 ft² is about .093 m². An rx-8 is about 52.8 inches tall and 69.7 inches wide. I assumed it took about 80% of that area up. That's 25.56 ft² or 1.90 m².
DragPower = .5*.31*82.7³*1.9*1.22
= 203157.2 watts
~ = 272.3 whp (1 hp, depending on definition, is about 746 watts)
Edit: Correcting for my left out variable, I again get 270 whp required which agrees with my initial post. In general, I try not to just pull **** out of my ***.
Edit: Took out comments about the incorrect calculation as this is now representative of how I did it.
Last edited by maxxdamigz; 05-11-2007 at 11:56 AM.
05-11-2007, 10:29 AM
#90
WIKI-WIKI-WIKI:
The power required to overcome the aerodynamic drag is given by:
where
V is the velocity of the object.
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a work in half the time requires eight times the power.
It should be emphasized here that the drag equation is an approximation, and does not necessarily give a close approximation in every instance. Thus one should be careful when making assumptions using these equations.
Drag equation:
where
Fd is the force of drag,
ρ is the density of the fluid (Note that for the Earth's atmosphere, the density can be found using the barometric formula. It is 1.293 kg/m3 at 0°C and 1 atmosphere.),
v is the speed of the object relative to the fluid,
A is the reference area,
Cd is the drag coefficient (a dimensionless constant, e.g. 0.25 to 0.45 for a car), and
is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).
The reference area A is related to, but not exactly equal to, the area of the projection of the object on a plane perpendicular to the direction of motion (i.e., cross sectional area). Sometimes different reference areas are given for the same object in which case a drag coefficient corresponding to each of these different areas must be given. The reference for a wing would be the plane area rather than the frontal area.
The power required to overcome the aerodynamic drag is given by:
where
V is the velocity of the object.
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a work in half the time requires eight times the power.
It should be emphasized here that the drag equation is an approximation, and does not necessarily give a close approximation in every instance. Thus one should be careful when making assumptions using these equations.
Drag equation:
where
Fd is the force of drag,
ρ is the density of the fluid (Note that for the Earth's atmosphere, the density can be found using the barometric formula. It is 1.293 kg/m3 at 0°C and 1 atmosphere.),
v is the speed of the object relative to the fluid,
A is the reference area,
Cd is the drag coefficient (a dimensionless constant, e.g. 0.25 to 0.45 for a car), and
is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).The reference area A is related to, but not exactly equal to, the area of the projection of the object on a plane perpendicular to the direction of motion (i.e., cross sectional area). Sometimes different reference areas are given for the same object in which case a drag coefficient corresponding to each of these different areas must be given. The reference for a wing would be the plane area rather than the frontal area.
Last edited by eviltwinkie; 05-11-2007 at 02:18 PM.
05-11-2007, 11:43 AM
#91
I realized I forgot to add in density today. I will edit my post. It should work back out the way I initially reported.
The chief inaccuracies in my calc would stem from the following:
1. Variation of Cd based on aftermarket or OEM body part variation.
2. I approximated density with a temperature guess.
3. Frontal area approximation - this is a flat out guesstimate but not orders of magnitude off.
Additional WHP would be needed to overcome:
1. Skin friction based on material and finish. Aerodynamic drag is generally only shape based.
2. Rolling resistance - tires deflectwhen rolled etc. These are generally lumped into the drive train loss BHP -> WHP and I don't think anyone really wants to calculate them.
3. Head wind - conversely tail wind would decrease HP required. At 185 mph, strong side wind would probably be a bad thing.
The chief inaccuracies in my calc would stem from the following:
1. Variation of Cd based on aftermarket or OEM body part variation.
2. I approximated density with a temperature guess.
3. Frontal area approximation - this is a flat out guesstimate but not orders of magnitude off.
Additional WHP would be needed to overcome:
1. Skin friction based on material and finish. Aerodynamic drag is generally only shape based.
2. Rolling resistance - tires deflectwhen rolled etc. These are generally lumped into the drive train loss BHP -> WHP and I don't think anyone really wants to calculate them.
3. Head wind - conversely tail wind would decrease HP required. At 185 mph, strong side wind would probably be a bad thing.
Last edited by maxxdamigz; 05-11-2007 at 11:54 AM.
05-11-2007, 12:01 PM
#92
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you can't calculate a suck situation though... Top speed is very much affected by the environment. I can't hit 160 on a flat highway, but with an incline, I've hit 160 before... Only on the most pristine of days...
Also changing my tires from 225/45/18 too 245/40/18 somehow doesn't allow me too reach 160 ever...
Also changing my tires from 225/45/18 too 245/40/18 somehow doesn't allow me too reach 160 ever...
05-11-2007, 01:04 PM
#93
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where's Stallion with his video???
05-11-2007, 01:37 PM
#94
FORMULA: .5 * p * v^3 * A * Cd
165 MPH:
.5 * 1.204 * 73.755^3 * 1.90 * .31 = 142261.10839531950975
p = 1.204 (68F/20C @ 1 atm sea level with dry air - Dense cool air for max HP possible)
v = 73.755 (165 mph * .447 = m/s)
A = 1.90
Cd = .31 (According to Mazda)
Metric horsepower, as a rule, is defined as 0.73549875 kW, or roughly 98.6% of mechanical horsepower.
1hp = 735.49875 W
142261.10839531950975 / 735.49875 = 193.42127827589035297476712230986
CONCLUSION:
So under these calculated conditions you could hit 165 mph in the RX...
Not factoring in rolling resistance, weight, windspeed, or distance or time required and under absolutely perfect ideal conditions. You are at sea level (0 altitude) on an absolutely level road when its 68F outside and your car has 193whp.
185 MPH:
.5 * 1.204 * 82.695^3 * 1.90 * .31 = 200516.23462025542275
p = 1.204 (68F/20C @ 1 atm sea level with dry air - Dense cool air for max HP possible)
v = 82.695 (185 mph * .447 = m/s)
A = 1.90
Cd = .31 (According to Mazda)
Metric horsepower, as a rule, is defined as 0.73549875 kW, or roughly 98.6% of mechanical horsepower.
1hp = 735.49875 W
200516.23462025542275 / 735.49875 = 272.62620721008080945072986187944
CONCLUSION:
So under these calculated conditions you could hit 185 mph in the RX...
Not factoring in rolling resistance, weight, windspeed, or distance or time required and under absolutely perfect ideal conditions. You are at sea level (0 altitude) on an absolutely level road when its 68F outside and your car has 273whp.
165 MPH:
.5 * 1.204 * 73.755^3 * 1.90 * .31 = 142261.10839531950975
p = 1.204 (68F/20C @ 1 atm sea level with dry air - Dense cool air for max HP possible)
v = 73.755 (165 mph * .447 = m/s)
A = 1.90
Cd = .31 (According to Mazda)
Metric horsepower, as a rule, is defined as 0.73549875 kW, or roughly 98.6% of mechanical horsepower.
1hp = 735.49875 W
142261.10839531950975 / 735.49875 = 193.42127827589035297476712230986
CONCLUSION:
So under these calculated conditions you could hit 165 mph in the RX...
Not factoring in rolling resistance, weight, windspeed, or distance or time required and under absolutely perfect ideal conditions. You are at sea level (0 altitude) on an absolutely level road when its 68F outside and your car has 193whp.
185 MPH:
.5 * 1.204 * 82.695^3 * 1.90 * .31 = 200516.23462025542275
p = 1.204 (68F/20C @ 1 atm sea level with dry air - Dense cool air for max HP possible)
v = 82.695 (185 mph * .447 = m/s)
A = 1.90
Cd = .31 (According to Mazda)
Metric horsepower, as a rule, is defined as 0.73549875 kW, or roughly 98.6% of mechanical horsepower.
1hp = 735.49875 W
200516.23462025542275 / 735.49875 = 272.62620721008080945072986187944
CONCLUSION:
So under these calculated conditions you could hit 185 mph in the RX...
Not factoring in rolling resistance, weight, windspeed, or distance or time required and under absolutely perfect ideal conditions. You are at sea level (0 altitude) on an absolutely level road when its 68F outside and your car has 273whp.
Last edited by eviltwinkie; 05-11-2007 at 02:10 PM.
05-11-2007, 01:39 PM
#95
you can't calculate a suck situation though... Top speed is very much affected by the environment. I can't hit 160 on a flat highway, but with an incline, I've hit 160 before... Only on the most pristine of days...
Also changing my tires from 225/45/18 too 245/40/18 somehow doesn't allow me too reach 160 ever...
Also changing my tires from 225/45/18 too 245/40/18 somehow doesn't allow me too reach 160 ever...
05-11-2007, 02:06 PM
#97
Haaa...I knows it...however most Euro-peon / asian vendors all spec to the metric HP spec...at least thats what I was under the impression of. That said since we were talking about an asian car technically, I figured I would use metric HP. Results appear to come out close enough either way.
05-11-2007, 03:01 PM
#100
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Isn't going 165mph in the 8 a bad idea unless you've got high end rubber rated to do it? Also, it seems like approx. 150mph was the intended design limit of the car... do you need special mods to make downforce or is it enough to just lower your car? Anyone have any thoughts on this??