Power:Weight and Aerodynamics question
03-20-2004, 08:31 AM
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Power:Weight and Aerodynamics question
How much of a role does power to weight ratio play in terms of acceleration after, lets say, 100 mph? Or does the vehicle's aerodynamics alone determine its ability to achieve a higher speed. I know that frontal area is major player, especially at higher speeds. Can any of you physics guys chime in hear. I didn't pay attention in my science classes in h.s. 
03-20-2004, 10:44 AM
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Re: Power:Weight and Aerodynamics question (Rah Rah)
Example:2 F up the other one nice catch.
A Honda Civic that weighs 2200lbs. Accelerating a .500g at 100MPH
2200lbs. X (.500g + .075g ) X 100MPH / 375 = 337.33 HP (gross)
2200lbs. x .075 x 100MPH / 375 = 44 HP (gross)
So if the civic was under an acceleration of .500g at 100MPH, the gross HP would be 337HP, and of that 337HP, 44HP are used to overcome vehicular drag and 293HP are used to accelerate the vehicle.
g=(gravity units)
(frictional drag) + (viscous drag) + (aero. drag)
.015 + (.0001 x 100MPH) + (.000005 x 100square) = .075g
Deceleration due to frictional drag is approximately .015g at all speeds.
Deceleration due to viscous drag is approximately .0001g x MPH.
Deceleration due to aerodynamics drag is approximately .000005g x MPH square
This formula kinda gives you a understanding of whats going on by changing some of the inputs except 375 its a constant.
Modified by RA166E at 7:04 AM 3/21/2004
Modified by RA166E at 7:08 AM 3/21/2004
A Honda Civic that weighs 2200lbs. Accelerating a .500g at 100MPH
2200lbs. X (.500g + .075g ) X 100MPH / 375 = 337.33 HP (gross)
2200lbs. x .075 x 100MPH / 375 = 44 HP (gross)
So if the civic was under an acceleration of .500g at 100MPH, the gross HP would be 337HP, and of that 337HP, 44HP are used to overcome vehicular drag and 293HP are used to accelerate the vehicle.
g=(gravity units)
(frictional drag) + (viscous drag) + (aero. drag)
.015 + (.0001 x 100MPH) + (.000005 x 100square) = .075g
Deceleration due to frictional drag is approximately .015g at all speeds.
Deceleration due to viscous drag is approximately .0001g x MPH.
Deceleration due to aerodynamics drag is approximately .000005g x MPH square
This formula kinda gives you a understanding of whats going on by changing some of the inputs except 375 its a constant.
Modified by RA166E at 7:04 AM 3/21/2004
Modified by RA166E at 7:08 AM 3/21/2004
03-20-2004, 12:37 PM
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Re: Power:Weight and Aerodynamics question (RA166E)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RA166E »</TD></TR><TR><TD CLASS="quote">Example:
A Honda Civic that weighs 2200lbs. Accelerating a .500g at 100MPH
2200lbs. X (.500g + .043g ) X 100MPH / 375 = 318.56 HP (gross)
The .043g is from the average vehicular drag characteristics.
(frictional drag) + (viscous drag) + (aero. drag)
This formula kinda gives you a understanding of whats going on by changing some of the inputs except 375 its a constant.</TD></TR></TABLE>
Could you please site a source for that .043g figure and give correct units? I'm almost sure that if you let off the gas and coasted in neutral at 100mph that you would decelerate at about twice that.
To answer the original question, the acceleration is still all about power to weight, but it is the power left after overcoming aerodynamic and frictional drag! Take you wheel hp at that speed and subtract the power needed to overcome aero drag and rolling resistance. An approximate formula for finding the drag force for your car at a certain speed is Fd = .5 * Cd * p * V^2 * A , where Fd is the drag force, Cd is the coefficient of drag (~.30-.35 for a stock Civic, ~.35-.40 if you rice it up w/body kit, wing, phat tires, etc), p (the Greek letter rho) = 0.00238 slug/ft^3 is the density of sea level air, V = velocity, and A = the frontal area of your car. DO NOT forget to cancel units with (slug*ft)/(lbf*s^2). The rolling resistance = Ihavenoidea. 1 hp = 550 ft*lbf/s. 100 mph = 147 ft/s. Also, aerodynamic drag is very nearly a function of the SQUARE of the velocity, so the drag at 100mph is four times the drag at 50mph. If you know it at one speed then you can closely estimate it at any other speed.
What hp is left over is what you will use to calculate your acceleration from f=ma, where f is the force from the contact patch of your tire, i.e. engine torque at that rpm x tranny and diff gearing x drivetrain loss factor (~.90, 10% loss?) / length of lever arm (axle centerline to contact patch). ALWAYS use proper units when writing this stuff down or you'll be sorry. Torque = hp x 5250/rpm.
A Honda Civic that weighs 2200lbs. Accelerating a .500g at 100MPH
2200lbs. X (.500g + .043g ) X 100MPH / 375 = 318.56 HP (gross)
The .043g is from the average vehicular drag characteristics.
(frictional drag) + (viscous drag) + (aero. drag)
This formula kinda gives you a understanding of whats going on by changing some of the inputs except 375 its a constant.</TD></TR></TABLE>
Could you please site a source for that .043g figure and give correct units? I'm almost sure that if you let off the gas and coasted in neutral at 100mph that you would decelerate at about twice that.
To answer the original question, the acceleration is still all about power to weight, but it is the power left after overcoming aerodynamic and frictional drag! Take you wheel hp at that speed and subtract the power needed to overcome aero drag and rolling resistance. An approximate formula for finding the drag force for your car at a certain speed is Fd = .5 * Cd * p * V^2 * A , where Fd is the drag force, Cd is the coefficient of drag (~.30-.35 for a stock Civic, ~.35-.40 if you rice it up w/body kit, wing, phat tires, etc), p (the Greek letter rho) = 0.00238 slug/ft^3 is the density of sea level air, V = velocity, and A = the frontal area of your car. DO NOT forget to cancel units with (slug*ft)/(lbf*s^2). The rolling resistance = Ihavenoidea. 1 hp = 550 ft*lbf/s. 100 mph = 147 ft/s. Also, aerodynamic drag is very nearly a function of the SQUARE of the velocity, so the drag at 100mph is four times the drag at 50mph. If you know it at one speed then you can closely estimate it at any other speed.
What hp is left over is what you will use to calculate your acceleration from f=ma, where f is the force from the contact patch of your tire, i.e. engine torque at that rpm x tranny and diff gearing x drivetrain loss factor (~.90, 10% loss?) / length of lever arm (axle centerline to contact patch). ALWAYS use proper units when writing this stuff down or you'll be sorry. Torque = hp x 5250/rpm.
03-21-2004, 09:33 PM
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Something that is interesting to note is that top speed is that weight has no bearing at all. Top speed is 100% compromised of aerodynamics and power.
Also, note that acceleration decreases in higher gears. This is not just due to aero drag, but also due to the fact that wider gear ratios give you less thrust.
Also, note that acceleration decreases in higher gears. This is not just due to aero drag, but also due to the fact that wider gear ratios give you less thrust.
03-22-2004, 12:14 AM
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Re: Power:Weight and Aerodynamics question (93LSivic)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 93LSivic »</TD></TR><TR><TD CLASS="quote">I dont know about all that. Just turn the boost up 
</TD></TR></TABLE>
you would be the one to say that
</TD></TR></TABLE>you would be the one to say that
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03-22-2004, 04:25 PM
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Re: Power:Weight and Aerodynamics question (Rah Rah)
Aerodynamics start playing a major role at 45mph in most cars.
Take my yamaha R6 for example. It can obliterate most cars to 100mph. The only problem is that there is only ~100 hp pushing me and the bike through the wind in triple digit speeds, and thus I am limited to a realistic (non speedo reading) 160mph or so top speed.
My eclipse can easilly reach or exceede 160mph. However, the eclipse stands no chance at all against the bike from a 30mph roll.
Equations are great, but most people think of things in a static world. Horsepower is not constant vs. rpm. Acceleration is not constant with respect to the gear you are in, you have the force of air resistance working against you in a linear fashon. Rolling resistance of the tires increases with speed (minor toe in). Wheel bearings become less efficient at higher speeds, the same is true of transmissions and general driveline equipment. Ride height of the car changes if the car produces downforce at speed (which most cars do). The temperature of the air entering the engine changes with speed and alters power output.
What it all comes down to however is what was explained above. Air resistance is the dominat force limiting your top speed with a given power output.
I get to study this s*** in class. Higher order Ordinary Differential Equations. We are using a spring - mass system right now, but the principles are simillar. In some situations, some variables (such as vehicle weight at speed) cancle out.
Take my yamaha R6 for example. It can obliterate most cars to 100mph. The only problem is that there is only ~100 hp pushing me and the bike through the wind in triple digit speeds, and thus I am limited to a realistic (non speedo reading) 160mph or so top speed.
My eclipse can easilly reach or exceede 160mph. However, the eclipse stands no chance at all against the bike from a 30mph roll.
Equations are great, but most people think of things in a static world. Horsepower is not constant vs. rpm. Acceleration is not constant with respect to the gear you are in, you have the force of air resistance working against you in a linear fashon. Rolling resistance of the tires increases with speed (minor toe in). Wheel bearings become less efficient at higher speeds, the same is true of transmissions and general driveline equipment. Ride height of the car changes if the car produces downforce at speed (which most cars do). The temperature of the air entering the engine changes with speed and alters power output.
What it all comes down to however is what was explained above. Air resistance is the dominat force limiting your top speed with a given power output.
I get to study this s*** in class. Higher order Ordinary Differential Equations. We are using a spring - mass system right now, but the principles are simillar. In some situations, some variables (such as vehicle weight at speed) cancle out.
03-22-2004, 05:31 PM
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Re: (kpt4321)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by kpt4321 »</TD></TR><TR><TD CLASS="quote">Something that is interesting to note is that top speed is that weight has no bearing at all. Top speed is 100% compromised of aerodynamics and power.
Also, note that acceleration decreases in higher gears. This is not just due to aero drag, but also due to the fact that wider gear ratios give you less thrust.</TD></TR></TABLE>
This why the fastest car on the planet are like 6000lbs. Just to carry the engines needed, and the fuel.
Danl how far have you gotten into those equations? I never had to do those, and curious if I could get a few from you. I like doing that stuff.
Also, note that acceleration decreases in higher gears. This is not just due to aero drag, but also due to the fact that wider gear ratios give you less thrust.</TD></TR></TABLE>
This why the fastest car on the planet are like 6000lbs. Just to carry the engines needed, and the fuel.
Danl how far have you gotten into those equations? I never had to do those, and curious if I could get a few from you.
I like doing that stuff.
03-22-2004, 05:43 PM
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Re: (nd_styles)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nd_styles »</TD></TR><TR><TD CLASS="quote">Danl how far have you gotten into those equations? I never had to do those, and curious if I could get a few from you. I like doing that stuff.</TD></TR></TABLE>
Like it? You're sick! I left my Diffy Q's class behind without ever looking back!
I like doing that stuff.</TD></TR></TABLE>Like it? You're sick! I left my Diffy Q's class behind without ever looking back!
03-22-2004, 05:49 PM
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Re: (nd_styles)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nd_styles »</TD></TR><TR><TD CLASS="quote">
Danl how far have you gotten into those equations? I never had to do those, and curious if I could get a few from you. I like doing that stuff.</TD></TR></TABLE>
We've been talking about it for a few days and starting analyzing what happens when you swing each variable each way. Their are a bunch of ODE equations online that explain things with respect to air resistance. A straight linear ODE (1st order I belive) is what we used to calculate the terminal velocity of an object. Something simillar would be used to model a cars top speed. Only difference is the force is not linear as the engine power changes with rpm. It could be modeled, and i'm sure you could get resonably close, plus or minus 5mph I'd say if you have an accurate coefficient of drag. You can add in all the variables you would want to account for. Just sum all the forces in the system that you want to account for just like you would in physics class. Solve the ODE then plug in t----->infinity. Thing is, how do you solve such an ugly ODE?
Engine power changing with RPM has got me baffled. The equation we used in class was with an object falling with a constant of gravitiy being the force. Since engine power is non-linear (IE: not a constant), the equation all of a sudden gets very very ugly
I'm far from an expert, hardly an amateur even. If you like ODE's then their is a ton of published research redially available to you to go have fun with.
Modified by danl at 7:05 PM 3/22/2004
Danl how far have you gotten into those equations? I never had to do those, and curious if I could get a few from you.
I like doing that stuff.</TD></TR></TABLE>We've been talking about it for a few days and starting analyzing what happens when you swing each variable each way. Their are a bunch of ODE equations online that explain things with respect to air resistance. A straight linear ODE (1st order I belive) is what we used to calculate the terminal velocity of an object. Something simillar would be used to model a cars top speed. Only difference is the force is not linear as the engine power changes with rpm. It could be modeled, and i'm sure you could get resonably close, plus or minus 5mph I'd say if you have an accurate coefficient of drag. You can add in all the variables you would want to account for. Just sum all the forces in the system that you want to account for just like you would in physics class. Solve the ODE then plug in t----->infinity. Thing is, how do you solve such an ugly ODE?
Engine power changing with RPM has got me baffled. The equation we used in class was with an object falling with a constant of gravitiy being the force. Since engine power is non-linear (IE: not a constant), the equation all of a sudden gets very very ugly
I'm far from an expert, hardly an amateur even. If you like ODE's then their is a ton of published research redially available to you to go have fun with.
Modified by danl at 7:05 PM 3/22/2004
03-22-2004, 06:18 PM
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Re: (danl)
DaX - I'm a physics major. I have to be crazy, and I love math. 
Just me.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by danl »</TD></TR><TR><TD CLASS="quote">
We've been talking about it for a few days and starting analyzing what happens when you swing each variable each way. Their are a bunch of ODE equations online that explain things with respect to air resistance. A straight linear ODE (1st order I belive) is what we used to calculate the terminal velocity of an object. Something simillar would be used to model a cars top speed. Only difference is the force is not linear as the engine power changes with rpm. It could be modeled, and i'm sure you could get resonably close, plus or minus 5mph I'd say if you have an accurate coefficient of drag. You can add in all the variables you would want to account for. Just sum all the forces in the system that you want to account for just like you would in physics class. Solve the ODE then plug in t----->infinity. Thing is, how do you solve such an ugly ODE?
Engine power changing with RPM has got me baffled. The equation we used in class was with an object falling with a constant of gravitiy being the force. Since engine power is non-linear (IE: not a constant), the equation all of a sudden gets very very ugly
I'm far from an expert, hardly an amateur even. If you like ODE's then their is a ton of published research redially available to you to go have fun with.
</TD></TR></TABLE>
Any site's with these equations? I was looking into a program that I could use to scan my car into, and figure out the drag co-efficient, and try to improve it's current aero situation. I was hopinh I could find a cheap one, but that's not working so far. I really don't want to try and calculate the whole car, but if I have to I will.
Just me. <TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by danl »</TD></TR><TR><TD CLASS="quote">
We've been talking about it for a few days and starting analyzing what happens when you swing each variable each way. Their are a bunch of ODE equations online that explain things with respect to air resistance. A straight linear ODE (1st order I belive) is what we used to calculate the terminal velocity of an object. Something simillar would be used to model a cars top speed. Only difference is the force is not linear as the engine power changes with rpm. It could be modeled, and i'm sure you could get resonably close, plus or minus 5mph I'd say if you have an accurate coefficient of drag. You can add in all the variables you would want to account for. Just sum all the forces in the system that you want to account for just like you would in physics class. Solve the ODE then plug in t----->infinity. Thing is, how do you solve such an ugly ODE?
Engine power changing with RPM has got me baffled. The equation we used in class was with an object falling with a constant of gravitiy being the force. Since engine power is non-linear (IE: not a constant), the equation all of a sudden gets very very ugly
I'm far from an expert, hardly an amateur even. If you like ODE's then their is a ton of published research redially available to you to go have fun with.
</TD></TR></TABLE>
Any site's with these equations? I was looking into a program that I could use to scan my car into, and figure out the drag co-efficient, and try to improve it's current aero situation. I was hopinh I could find a cheap one, but that's not working so far. I really don't want to try and calculate the whole car, but if I have to I will.
03-22-2004, 06:28 PM
#15
Engine power changing with engine speed will be a second or third order DE, and maybe non-homogeneous. I haven't taken diffy q's in two years, and really all I use from that class is laplace transforms.
Physics major eh nd_styles? ME major here...I took Modern Physics as a science elective and dropped it after the first week. Physics 1&2 and that's all I want of that!
Physics major eh nd_styles? ME major here...I took Modern Physics as a science elective and dropped it after the first week. Physics 1&2 and that's all I want of that!
03-22-2004, 10:39 PM
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This thread just got wicked freaking cool.
I'm a ME Major as well, and I actually just finished up a differential equations class last term.
Engine power versus speed is not going to be easy to model. You could whip up an approximation, but it would probably still be a pretty nasty polynomial. I'd think at least third order, perhaps worse.
Of course, it depends on the vehicle as well.
I'm a ME Major as well, and I actually just finished up a differential equations class last term.
Engine power versus speed is not going to be easy to model. You could whip up an approximation, but it would probably still be a pretty nasty polynomial. I'd think at least third order, perhaps worse.
Of course, it depends on the vehicle as well.
03-23-2004, 03:23 AM
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Re: (kpt4321)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by kpt4321 »</TD></TR><TR><TD CLASS="quote">This thread just got wicked freaking cool.
I'm a ME Major as well, and I actually just finished up a differential equations class last term.
Engine power versus speed is not going to be easy to model. You could whip up an approximation, but it would probably still be a pretty nasty polynomial. I'd think at least third order, perhaps worse.
</TD></TR></TABLE>
Yeah I know. The more I think about modeling this situation the uglier it gets. Like I said, when we modeled terminal velocity with force being a constant, that was a 1st order non-homogenous equation. I don't even want to do it for when engine power is dynamic, IF it can even be done.
I'm a ME Major as well, and I actually just finished up a differential equations class last term.
Engine power versus speed is not going to be easy to model. You could whip up an approximation, but it would probably still be a pretty nasty polynomial. I'd think at least third order, perhaps worse.
</TD></TR></TABLE>
Yeah I know. The more I think about modeling this situation the uglier it gets. Like I said, when we modeled terminal velocity with force being a constant, that was a 1st order non-homogenous equation. I don't even want to do it for when engine power is dynamic, IF it can even be done.
03-23-2004, 05:56 AM
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Re: (Rah Rah)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Rah Rah »</TD></TR><TR><TD CLASS="quote">I guess what I"m asking is if weight plays any factor in hindering acceleration at high speeds. </TD></TR></TABLE>
Yes, just as it does at any other speed, only at high speeds you have less and less power to accelerate it with. It won't effect top speed, but more weight will make it take longer to get there because you are accelerating more slowly.
Modified by tjbizzo at 9:07 AM 3/24/2004
Yes, just as it does at any other speed, only at high speeds you have less and less power to accelerate it with. It won't effect top speed, but more weight will make it take longer to get there because you are accelerating more slowly.
Modified by tjbizzo at 9:07 AM 3/24/2004
03-23-2004, 09:32 AM
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The slower your rate of acceleration (dv/dt, for the sake of calculus), then the less of an effect weight has on the car. This is why weight does not effect top speed, as your acceleration at top speed is obviously zero.
), then the less of an effect weight has on the car. This is why weight does not effect top speed, as your acceleration at top speed is obviously zero.
03-23-2004, 09:37 AM
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Re: (danl)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by danl »</TD></TR><TR><TD CLASS="quote">
Yeah I know. The more I think about modeling this situation the uglier it gets. Like I said, when we modeled terminal velocity with force being a constant, that was a 1st order non-homogenous equation. I don't even want to do it for when engine power is dynamic, IF it can even be done.</TD></TR></TABLE>
There is another way to do it, which is quite a bit easier.
First, you need a dyno chart of your car. Then, you can use the gear ratios and tire diameter to find the thrust (force) in your highest gear, and graph that out. It will follow the torque curve, to some extent. Then, using some aero information about your car, you can find the drag compared to speed. If you graph out drag as force, and thrust as force, and index them both by ground speed, then you can find the point where the two curve cross. This point is where the required thrust goes above the thrust provided, and where the car stops accelerating.
Yeah I know. The more I think about modeling this situation the uglier it gets. Like I said, when we modeled terminal velocity with force being a constant, that was a 1st order non-homogenous equation. I don't even want to do it for when engine power is dynamic, IF it can even be done.</TD></TR></TABLE>
There is another way to do it, which is quite a bit easier.
First, you need a dyno chart of your car. Then, you can use the gear ratios and tire diameter to find the thrust (force) in your highest gear, and graph that out. It will follow the torque curve, to some extent. Then, using some aero information about your car, you can find the drag compared to speed. If you graph out drag as force, and thrust as force, and index them both by ground speed, then you can find the point where the two curve cross. This point is where the required thrust goes above the thrust provided, and where the car stops accelerating.
03-24-2004, 02:56 AM
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Re: (kpt4321)
Thats exactly right ha ha. Don't work harder, work smarter.
I used that method when I was trying to figure out how far different fuel pumps would take me if injector flow was held fixed, but fuel pressure changed (raising boost pressure raises fuel pressure). Changing fuel pressure affects the fuel pump flow (which most people don't compensate for).
FWIW, a 190lph pump can support 650cc injectors at 90% duty cycle up until 60psi fuel rail pressure. Pressure should be measured at the fuel pump not the fuel rail.
I used that method when I was trying to figure out how far different fuel pumps would take me if injector flow was held fixed, but fuel pressure changed (raising boost pressure raises fuel pressure). Changing fuel pressure affects the fuel pump flow (which most people don't compensate for).
FWIW, a 190lph pump can support 650cc injectors at 90% duty cycle up until 60psi fuel rail pressure. Pressure should be measured at the fuel pump not the fuel rail.
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