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Discussion Starter #1
Das,

Okay, here we go, in its simplest form:

Air density

0.076 psi @ 60°F
0.068 psi @ 120°F

So the air is 10% less dense @ 120°F, requiring 10% less horsepower to do the same speed.

For the same example, the SAE correction factor shows a HP drop of 5% from 60°F to 120°F.

From that, you can conlude that you actually have a 5% power advantage over running in the lower temps.

I know it's an extreme example, but it's one that gives you a good idea of what really goes on.

I can't find all the hardcore math we did for one of our projects similar to this, but hopefully i can find them. I'd rather not derive it all again.

In our project we assumed runs from sea level, but i think SCCs example was just assuming that bot hruns occurred at the same altitude to keep everything consistent.

I'd dole out more rep if i could, good discussion!!!
 

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Seriously... 10% less dense = 10% more power? Are you kidding me?

http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

I'm pretty sure it's not a linear relationship. Newtons second law is a direct relationship, but the aerodynamic drag varies based on the square of velocity.

Ugh... am I going to have to dust off the calculus? :( We're going to need some science people up in here.

Lots of reading to do if we want to go down this path:

Earth Atmosphere Model
http://www.grc.nasa.gov/WWW/K-12/airplane/atmos.html

Properties of Air
http://www.grc.nasa.gov/WWW/K-12/airplane/airprop.html

+rep to anyone who can find an equation that takes in force, air density, and drag and spits out accelleration.
 

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in real world applications it's a bit different because you have to factor in humidity and other variables as well...

we had a customer once who lived on the big island (hawaii) and he would drive up to the observatory near the top of a volcano there...the air is so thin up there that he wanted to buy a supercharger to save on gas. we thought that was weird but he showed us his calcuations...i wish i wrote them down but it was pretty nuts (he was some sort of scientist)

anyway...good discussion... :)
 

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Here's what I have for now:

drag force = ½ρV2 × cd× A

p : pressure
V : velocity
cd : coefficient of drag
A : area


a = m/F
a : accelleration
m : mass
F = force


And for that, F would be:

Thrust (horsepower?) - drag = F

So you'd end up with this I think:

Accelleration = mass / ((force or horsepower) - (½ρV2 × cd× A))


Of course, you'd have to keep all your units straight, and find out how the cd for the IS300 was calculated by Lexus (I assume it takes into account "constants" such as air temp and humidity, which we'd have to unravel to compute different cd as well as p pressures.

Seriously, we need some sort of Math genious. Anyone know one?
 

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Discussion Starter #5 (Edited)
I know the Aero HP equation, and your eq needs to be adjusted.

Aero HP = Frontal Area x N x Cd x speed^3

The value from this equation will be the RWHP needed to overcome air drag at the speed entered in the EQ.

N= .000007 assuming STP = .08

Assuming STP, you don't need to know how lexus figures out the Cd. But i can answer that for you: Windtunnel.

After that, we can get into fluid dynamics and all that associated thoery.

In the article, i think they are trying to say that even assuming a HP loss of 5%, it is 10% EASIER to travel through the same medium. Where they fault, which is the problem we ran into with our project but eventually figured out is that just because it is 10% easier to travel through that medium, it doesn't DIRECTLY translate into 10% more power (you have a given amount of hp, which is 5% less than what you measured at 60°, whatever the number may be).

It ALSO doesn't translate into the car being 10% faster either. All it is saying is that the same car run at 60° will invariably be SLOWER than the same car run at 120° assuming STP.

Das,

When i go to work tomorrow, i will plug the values into a predictor we have at work (www.altronicsinc.com). We do handheld/trailor/PC based weather stations that take into account humidity, temp, GOW, vapor pressure, Dp, density/oxygen altitude and pressure for drag racing teams.

However, that only predicts 1/4 times after a baseline (can be arbitrary) run is entered. It doesn't take into account Cd or frontal area, but both of these variables are considered insignificant in drag racing (to an extent).
 

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Inspar8r said:
I know the Aero HP equation, and your math is a bit off.

Aero HP = Frontal Area x N x Cd x speed^3

N= .000007 assuming STP = .08

Assuming STP, you don't need to know how lexus figures out the Cd. But i can answer that for you: Windtunnel.

After that, we can get into fluid dynamics and all that associated thoery.

In the article, i think they are trying to say that even assuming a HP loss of 5%, it is 10% EASIER to travel through the same medium. Where they fault, which is the problem we ran into with our project but eventually figured out is that just because it is 10% easier to travel through that medium, it doesn't DIRECTLY translate into 10% more power (you have a given amount of hp, which is 5% less than what you measured at 60°, whatever the number may be).

It ALSO doesn't translate into the car being 10% faster either. All it is saying is that the same car run at 60° will invariably be SLOWER than the same car run at 120° assuming STP.
Going to have to cite your source on that, all I did was copy directly from tha NASA page and generalize.

I've gotten a few different equations for AeroHP from a google search:

F=1/2*Cd*A*rho*v^2
http://www.eng-tips.com/viewthread.cfm?qid=129383&page=1

Another...
http://www.eng-tips.com/viewthread.cfm?qid=129385&page=1

http://64.233.161.104/search?q=cach...+"Aero+HP"+-fork+-bike+-compaq+-hewlett&hl=en

Here's one with the cube as you stated:
http://www.clubcobra.com/t20273-15-6.html
AirDensity=0.0228645 * baro/(0.5555* (temp - 32) + 273.15)

AeroDragFactor = 1.07556 * AirDensity * cd * farea

RollingDragFactor = (C1 + C2 * mph ** (2.5))

C1 = 0.034667 * (tpress - 35.000000)*(tpress - 45.000000) +
-0.063333 * (tpress - 20.000000)*(tpress - 45.000000) +
0.032000 * (tpress - 20.000000)*(tpress - 35.000000);

C2 = 7.704e-07 * (tpress - 35.000000)*(tpress - 45.000000) +
-8.33333e-07 * (tpress - 20.000000)*(tpress - 45.000000) +
3.148e-07 * (tpress - 20.000000)*(tpress - 35.000000);

AeroHP = (AeroDragFactor * mph**3)/375.0

RollHP = RollingDragFactor * (weight/1000.) * (mph/375.0)

Lift = 0.00256 * farea * clift * mph**2

It lists aero hp as a function of the velocity cubed.
In any event... anything having to do with a power keeps it from being a linear relationship... I don't see the 10% coming from any of these equations... who's going to do the math? lol
 

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Here's some modelling software that says it can compute this answer for us...

http://tsrsoftware.com/maxspeed.htm

I sent an email to them asking if we could use it for our calculations. Will let you guys know if I get a reply.
 

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Discussion Starter #8
I'm not seeing where you are getting this 10% number. There isn't 10% more power. But in the example i gave you, it just says it is 10% easier (whether that is less power, or less torque, or less energy, i'm not sure).

SCC derived the EQ. I gave up trying to derive it to the version you found. I've had about enough derivations... LaPlace Transforms, MacLaurin and Fourier Series seem to have that kind of effect on everyone...

N is a variable that accounts for STP and whatnot. I'm not sure how they got that result.
 

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dasgalloway said:
Here's what I have for now:

drag force = ½ρV2 × cd× A

p : pressure
V : velocity
cd : coefficient of drag
A : area


a = m/F
a : accelleration
m : mass
F = force


And for that, F would be:

Thrust (horsepower?) - drag = F

So you'd end up with this I think:

Accelleration = mass / ((force or horsepower) - (½ρV2 × cd× A))


Of course, you'd have to keep all your units straight, and find out how the cd for the IS300 was calculated by Lexus (I assume it takes into account "constants" such as air temp and humidity, which we'd have to unravel to compute different cd as well as p pressures.

Seriously, we need some sort of Math genious. Anyone know one?
Das Thrust usually measures in ft/lbs ie torque

Ok I am trying to come up with the formula you requested (I am no math genuis but calculus III taught me a few things). But you confused me for a sec, figured it out now though....the P in your first equation is rho I believe and that is density.
 

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Inspar8r said:
I know the Aero HP equation, and your eq needs to be adjusted.

Aero HP = Frontal Area x N x Cd x speed^3

The value from this equation will be the RWHP needed to overcome air drag at the speed entered in the EQ.

N= .000007 assuming STP = .08

Assuming STP, you don't need to know how lexus figures out the Cd. But i can answer that for you: Windtunnel.

After that, we can get into fluid dynamics and all that associated thoery.

In the article, i think they are trying to say that even assuming a HP loss of 5%, it is 10% EASIER to travel through the same medium. Where they fault, which is the problem we ran into with our project but eventually figured out is that just because it is 10% easier to travel through that medium, it doesn't DIRECTLY translate into 10% more power (you have a given amount of hp, which is 5% less than what you measured at 60°, whatever the number may be).

It ALSO doesn't translate into the car being 10% faster either. All it is saying is that the same car run at 60° will invariably be SLOWER than the same car run at 120° assuming STP.

Das,

When i go to work tomorrow, i will plug the values into a predictor we have at work (www.altronicsinc.com). We do handheld/trailor/PC based weather stations that take into account humidity, temp, GOW, vapor pressure, Dp, density/oxygen altitude and pressure for drag racing teams.

However, that only predicts 1/4 times after a baseline (can be arbitrary) run is entered. It doesn't take into account Cd or frontal area, but both of these variables are considered insignificant in drag racing (to an extent).
Ok I just came up on this, but how are you assuming STP in both cases when the temp will change from 60-120 deg?? Also how do you assume STP again when you are looking at an area with less density vs an area with a higher density of air? When density is changed that way, you better believe in the real world that is that pressure will change and that will affect your STP assumption, by how much.......well I dont know that but it will
 

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kponti said:
Das Thrust usually measures in ft/lbs ie torque
not neccessarilly - thrust is generally in lbs (at least in my world(jet engine world)) the overall thrust of the vehicle will be in pounds as the torque emitted on teh wheel will result in a force vector on the very simplified scale so you have a net force of X which will be in # if it is to accelerate. acceleration is infact f=ma which is in #

torque will be on the wheel yes but that translates in a tangential force on teh road which is turned into thrust- but the car going forward is in pounds.

i see where you are comnig from but torque and thrust are very very differnet
 

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Inspar8r said:
N is a variable that accounts for STP and whatnot. I'm not sure how they got that result.
Inspar8r... do you know what STP is? Isn't it silly to vary temperature/pressure while keeping STP constant? :suspiciou

And you've got two forces that keep the car from going faster... inertia, and aerodynamic drag. Assuming inertia will stay the same at whatever temperature, it's the drag that's going to vary, and given that drag varies by velocity^2 and other factors, I don't see it having 10% less drag based on that temp change.
 

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somguy2u said:
not neccessarilly - thrust is generally in lbs (at least in my world(jet engine world)) the overall thrust of the vehicle will be in pounds as the torque emitted on teh wheel will result in a force vector on the very simplified scale so you have a net force of X which will be in # if it is to accelerate. acceleration is infact f=ma which is in #

torque will be on the wheel yes but that translates in a tangential force on teh road which is turned into thrust- but the car going forward is in pounds.

i see where you are comnig from but torque and thrust are very very differnet
Understood, but thrust is mearly a word used to describe a force which in vehicles is ft/lbs and in airplanes is the ability of the aircraft to move forward against weight, lift and drag.

How Does an Airplane Produce Thrust?

Thrust is the force created by a power source that overcomes the airplane's aerodynamic drag (its resistance to passing through the air) and gives it forward motion. This force can either "pull" or "push" the aircraft forward, depending on the type of power source used. Common types include reciprocating (piston-powered) engines driving propellers, and jet engines.
http://www.aeromuseum.org/Education/Lessons/HowPlaneFly/HowPlaneFly.html
 

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Discussion Starter #14
dasgalloway said:
Inspar8r... do you know what STP is? Isn't it silly to vary temperature/pressure while keeping STP constant? :suspiciou

And you've got two forces that keep the car from going faster... inertia, and aerodynamic drag. Assuming inertia will stay the same at whatever temperature, it's the drag that's going to vary, and given that drag varies by velocity^2 and other factors, I don't see it having 10% less drag based on that temp change.
STP = Standard Temerature and Pressure (.08).

It is what is used to determine Cd. Don't ask me why it is assumed. It's like asking why Pi = 3.14~ or why Avogadro's number = 6.022 x 10^23. SImilarly, it's like room temperature = 25°C.

They only used STP for calculating N. N is there to make sure the units all come out correct when you use the equation. Furthermore, I don't think using given temps and and pressure will vary enough to sway the result of N. Though I'm only making assumptions here. I don't know how they came up with N. According to the article, it is very complex and they had engineers do the math.
 

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STP or standard temperature and pressure, standard conditions for measurement of the properties of matter. The standard temperature is the freezing point of pure water, 0°C or 273.15K. The standard pressure is the pressure exerted by a column of mercury (symbol Hg) 760 mm high, often designated 760 mm Hg. This pressure is also called one atmosphere and is equal to 1.01325×106 dynes per sq cm, or approximately 14.7 lb per sq in. The density (mass per volume) of a gas is usually reported as its value at STP. Properties that cannot be measured at STP are measured under other conditions; usually the values obtained are then mathematically extrapolated to their values at STP.
 

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Discussion Starter #16
somguy2u said:
STP or standard temperature and pressure, standard conditions for measurement of the properties of matter. The standard temperature is the freezing point of pure water, 0°C or 273.15K. The standard pressure is the pressure exerted by a column of mercury (symbol Hg) 760 mm high, often designated 760 mm Hg. This pressure is also called one atmosphere and is equal to 1.01325×106 dynes per sq cm, or approximately 14.7 lb per sq in. The density (mass per volume) of a gas is usually reported as its value at STP. Properties that cannot be measured at STP are measured under other conditions; usually the values obtained are then mathematically extrapolated to their values at STP.
Much better explanation than mine ;)
 

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kponti said:
Understood, but thrust is mearly a word used to describe a force which in vehicles is ft/lbs and in airplanes is the ability of the aircraft to move forward against weight, lift and drag.


http://www.aeromuseum.org/Education/Lessons/HowPlaneFly/HowPlaneFly.html

I think the difference is syntactical... torque is in ft/lb's... and thrust is in lbs... units are different, so they must be describing different things.

I agree with someguy... applying torque over the wheel lever arm would probably get you to pure lbs of thrust in the end. In aero terms they don't use ft/lbs since there is no wheel/lever arm to work with... only pure thrust, which can be measured directly in lbs.

Can solve that part of the equation when we get there. Anyone know how to covert ft/lbs directly into lbs?
 

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dasgalloway said:
I think the difference is syntactical... torque is in ft/lb's... and thrust is in lbs... units are different, so they must be describing different things.

I agree with someguy... applying torque over the wheel lever arm would probably get you to pure lbs of thrust in the end. In aero terms they don't use ft/lbs since there is no wheel/lever arm to work with... only pure thrust, which can be measured directly in lbs.

Can solve that part of the equation when we get there. Anyone know how to covert ft/lbs directly into lbs?

you need to divide it by the moment arm length which is teh radius of the wheel generally (with tire)= lbs
 

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Inspar8r said:
STP = Standard Temerature and Pressure (.08).

It is what is used to determine Cd. Don't ask me why it is assumed. It's like asking why Pi = 3.14~ or why Avogadro's number = 6.022 x 10^23. SImilarly, it's like room temperature = 25°C.

They only used STP for calculating N. N is there to make sure the units all come out correct when you use the equation. Furthermore, I don't think using given temps and and pressure will vary enough to sway the result of N. Though I'm only making assumptions here. I don't know how they came up with N. According to the article, it is very complex and they had engineers do the math.
If it's used to determine Cd, and it takes temp and pressure into account, then you have to admit that Cd would vary when you moved away from STP.

Thats the entire point of the discussion here, right? How does Thurst and drag vary over diff temps and pressures?

Thrust increases as the engine gets a more dense air charge.
Drag increases as air density increases

Question is, for given changes of temp and pressure, which one increases faster?
 

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dasgalloway said:
If it's used to determine Cd, and it takes temp and pressure into account, then you have to admit that Cd would vary when you moved away from STP.

Thats the entire point of the discussion here, right? How does Thurst and drag vary over diff temps and pressures?

Thrust increases as the engine gets a more dense air charge.
Drag increases as air density increases

Question is, for given changes of temp and pressure, which one increases faster?

I think we can guess at it:

drag force = ½ρV2 × cd× A

p : pressure
V : velocity
cd : coefficient of drag
A : area
Wait... or not... if Cd changes based on temp and pressure, and all we have is Lexus' empiracly tested value for Cd, supposedly done @ STP... we can't finish the equation since we can't know how Cd changes.

Bah!
 
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