# Coefficient of Drag & Frontal Area



## madderscience (Jun 28, 2008)

only on this forum


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## Woodsmith (Jun 5, 2008)

Frontal area is going to be straight forward, width x height based on measurements or manufacturer's spec.

Cd will probably be 1.0 or as near as dammit! 

Unless you are planning on a major power/speed increase none of it matters except in a hurricane.


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## Sunking (Aug 10, 2009)

Woodsmith said:


> Frontal area is going to be straight forward, width x height based on measurements or manufacturer's spec.
> 
> Cd will probably be 1.0 or as near as dammit!
> 
> Unless you are planning on a major power/speed increase none of it matters except in a hurricane.


Well 600 mph it matters using Cd = 1 and Frontal area of 14/ft takes 14 hp to get to 60 mph. 12 hp of that is to overcome Cd.


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## Woodsmith (Jun 5, 2008)

I wouldn't fancy a ride, at 600mph, in a golf cart!

Be great to see though.


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## dragonsgate (May 19, 2012)

Woodsmith said:


> I wouldn't fancy a ride, at 600mph, in a golf cart!
> 
> Be great to see though.


I will bring the popcorn.


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## Sunking (Aug 10, 2009)

Woodsmith said:


> I wouldn't fancy a ride, at 600mph, in a golf cart!
> 
> Be great to see though.


Dang bouncing keys. 60 mph


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## dragonsgate (May 19, 2012)

Sunking said:


> Dang bouncing keys. 60 mph


60mph is still a good clip for a stock suspensioned golf cart. During my travels back when the speed limit was dropped to 55 I heard a lot of grossing about having to go so slow. During one bitch session someone asked why 55? someone else who claimed to be an engineer in something or the other said that up 59mph wind resistance did not play a big part. After that it became critical. Something about 59 being the break speed.


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## Woodsmith (Jun 5, 2008)

Sunking said:


> Dang bouncing keys. 60 mph


Ahhh, disappointed now.
I was looking forward to 600mph in a golf cart.


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## Hollie Maea (Dec 9, 2009)

dragonsgate said:


> Something about 59 being the break speed.


There isn't really a "break speed". Energy required to overcome wind losses increase by the square of velocity. So driving at 60 requires about 1.2 times more energy to overcome wind than driving at 55.

More likely, they started in the fuzzy region where wind losses start to dominate other losses and then just picked a number that "felt right" to the committee.


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## Sunking (Aug 10, 2009)

dragonsgate said:


> 60mph is still a good clip for a stock suspensioned golf cart.


My new one is a Ford Think which has a pretty decent suspension. It is a legal LSV so the windshield is DOT and cannot be removed.But as you can see it is curved. My EZGO cart does 30 mph easily.



If the VIDEO LINK worked you could see the world record holder for the fastest stock golf cart. Only the Motor, controller, and AGM batteries were changed out all of which you can buy. 118 mph, 12,241 sec 1/4 mile.

https://www.youtube.com/watch?v=Z93QpgGJRDA


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## Duncan (Dec 8, 2008)

I thought Woody was being a bit hard with a suggestion of "1" for a drag coefficient
Looking at that photo it may be more than "1"
Looks nice - but not aerodynamic


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## GerhardRP (Nov 17, 2009)

Sunking said:


> Anyone know what Cd and Frontal Area of a golf cart is? Cannot seem to find anything on the web. My WAG is .5 and 14/ft2


I think you can get a good estimate of all the dynamic losses by doing a careful measurement of speed during a level-ground coast-down measurement starting from your maximum speed.


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## IamIan (Mar 29, 2009)

Sorry I have no data on actual CdA numbers for gold carts.

- - - - - - 



Hollie Maea said:


> Energy required to overcome wind losses increase by the square of velocity. So driving at 60 requires about 1.2 times more energy to overcome wind than driving at 55.


Aerodynamic *force* increases with the square .. power and energy go up much faster.










- - - - - 
Power increases with the cube of speed.

The Power of that Wind Resistance (Drag) is:









- - - - - 
Energy = Power x Time
Soo ... if you travel for the same amount of time .. energy for wind resistance goes up with the cube of the speed the same as power does.

- - - - - - 

And Rolling resistance also has a difference between force and power.











Crr = The Coefficient of rolling resistance.

N = The 'normal' force .. of the force acting perpendicular to the road... this force decreases up or down hill .. and increases going around a banked turn.

P = F x V

Soo your rolling resistance power increases linearly with increases in speed.

- - - - - -

All else being equal .. overly simplified :
Going from 55 MPH to 60 MPH is a ~9% increase in speed .. 
Which increases the rolling resistance power by ~9%.

And also increases the wind resistance power by ~30%.

... Depending on the vehicle specifics of course ...
In a simple case of , if at 55 MPH both wind and rolling resistance were both contributing equal % share of the combined total ... and no significant other types of drag .. That ~9% speed increase would require a ~39% increase in power output.


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## kennybobby (Aug 10, 2012)

*Re: energy and power*

Well he did say energy and not 'power',

and energy, or work done, is also Force times Distance, and since the aero force is velocity squared, then so is the energy.

As you know the kinetic energy of an object in motion is 1/2 * mass * velocity squared.


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## Hollie Maea (Dec 9, 2009)

IamIan said:


> Soo ... if you travel for the same amount of time .. energy for wind resistance goes up with the cube of the speed the same as power does.


But you don't travel for the same amount of time, do you? Because you are going faster.

If you double your speed, your required power (neglecting everything but wind resistance) increases by the cube--8 times. But you get to your destination in half the time. So the energy used (which is what matters, since we were discussing fuel economy) increases by 8/2 = 4 times, or square. Which, as Kennybobby points out, matches the expected equation for work.


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## IamIan (Mar 29, 2009)

kennybobby said:


> Well he did say energy and not 'power',
> 
> and energy, or work done, is also Force times Distance, and since the aero force is velocity squared, then so is the energy.
> 
> As you know the kinetic energy of an object in motion is 1/2 * mass * velocity squared.





Hollie Maea said:


> But you don't travel for the same amount of time, do you? Because you are going faster.
> 
> If you double your speed, your required power (neglecting everything but wind resistance) increases by the cube--8 times. But you get to your destination in half the time. So the energy used (which is what matters, since we were discussing fuel economy) increases by 8/2 = 4 times, or square. Which, as Kennybobby points out, matches the expected equation for work.


I would still recommend at a minimum to at least including both aero and rolling resistance ... even if we wanted to simplify.

Any ground vehicle operating in an atmosphere will increase both it's power and energy needs per unit time or unit distance .. faster than the square of the speed ... expecting it to only increase at the square of the speed rate .. will only lead to less accurate under-estimates .. which will result in engine/motor systems not having enough power to achieve the desired speed , or acceleration .. and fuel/battery energy supplies being too small to achieve the desired distances.


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## Hollie Maea (Dec 9, 2009)

IamIan said:


> I would still recommend at a minimum to at least including both aero and rolling resistance ... even if we wanted to simplify.
> 
> Any ground vehicle operating in an atmosphere will increase both it's power and energy needs per unit time or unit distance .. faster than the square of the speed ... expecting it to only increase at the square of the speed rate .. will only lead to less accurate under-estimates .. which will result in engine/motor systems not having enough power to achieve the desired speed , or acceleration .. and fuel/battery energy supplies being too small to achieve the desired distances.


Well, technically yes. Especially at low speeds, the linear aspect of rolling resistance is very important. But as speed increases, a square factor completely swamps a linear factor.

But yeah, better not use square for figuring out power, since that increases by cube.


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## mizlplix (May 1, 2011)

I spent a ton of hours in two full size wind tunnels. We were optimizing the way body panels were hung on a stockcar. Our minimum critical speed was around 45 MPH. where we started to gain significant drag.

#1 get the car close to the ground to divert the wind over not under.
#2 On a short track car, we would hang the panels back to front. The laps made it wider at the front.
#3 On a speedway car, we would lap the panels front to back on the left side and back to front on the right side. The slightly crooked body liked to turn left easier.

The most important factor was to stop as much air from going under the car and also keep it out of the wheel wells (Fenders). Side skirts worked well at this.

Miz


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