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E&J push'n wind

San Diego CA, The best climate on earth!

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Posted: 04/10/12 01:54am Link  |  Print  |  Notify Moderator

Crunching numbers can be fun I supose for some people but in the end what does it matter? I'm sure the engineers have already done all the numbers right down to the tire compound matrix.

After you cook all the water off of it I'd have to agree with Old-biscuit!

But to answer your question.., well I can't! I aint no engineer. Oh.., maybe this thread is exclusively for engineers! OOPS.., sorry.


Wisdom is the right use of knowledge. To know is not to be wise. Many men know much, and are all the greater fools for it. There is no fool so great a fool as a knowing fool. But to know how to use knowledge is to have wisdom.

Charles Haddon Spurgeon


bucky

Eastern Shore of MD

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Posted: 04/10/12 04:23am Link  |  Print  |  Notify Moderator

Concrete or blacktop?


I just realized that my last 4 trucks were gas GMC, diesel Ford, gas Chevy, and now a diesel Dodge. I didn't realize I was so open minded lol.

NewsW

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Posted: 04/10/12 04:58am Link  |  Print  |  Notify Moderator

4x4ord wrote:

I was playing with the calculator and came up with some figures that seem to make some sense. Based on my experience with the truck I think the numbers are likely right but if there are some numbers guys out there that want to kill a little time I don't mind being corrected.

Based on:
11 mpg towing a 15,000 lb RV at 60mph on a level road with no wind;
and 18 HP hrs/gal (fairly optimistic efficiency for a diesel engine)
6.7 Powerstroke with 3.55 rear axle and 20" wheels
1.52:1 third gear and 2.32:1 2nd gear


These are the numbers I calculated:

HP required to overcome wind drag and rolling resistance and driveline loss is 98 HP on a level road with no wind.

Torque on the crankshaft at 60 mph is 365 ft lbs in 6th gear at 1411 rpm assuming level ground and no wind

lbs of pull between the drive tires and road surface is 490 lbs. (this assumes 20% power loss through the driveline).

Now start climbing a hill. A 7% grade and a 24,000 lb truck and trailer would put an additional 1676 lbs of pull on the surface of the drive tires.
This would equate to 335 hp at the crank (assuming 20% driveline loss)

So the total engine HP required to pull a 15000 lb trailer up a 7% grade at 60 mph would be very close to 433 HP.

To go down at 60 mph there would be the 1186 lbs of pull on the surface of the tires to overcome. This would create 249 ft lbs of torque on the crank assuming 3rd gear and 20% power absorbed into the driveline. To create that torque the exhaust brake would be required to hold 46 psi back pressure at 2920 rpm. The total power required to hold the unit back would be 139 HP from the exhaust brake plus 98 HP absorbed by the driveline, wind etc.

Now if you assume you go down at only 45 mph. The wind resistance would drop drastically so that the power holding the vehicle back due to wind, driveline and rolling resistance might be only 50 hp instead of 98. At 45 mph this equates to 417 lbs of pull on the outside surface of the rear wheels.

The hill and weight of the truck and trailer would still put 1676 lbs of pull at the surface of the pavement and rear wheels. So if you subtract the 417 that the wind and rolling resistance account for, and then account for the power absorbed by the driveline there would need to be 173 ft lbs of torque on the crank in order to hold the unit back from accelerating down the hill. (assuming the truck is shifted down into 2nd gear - 2.32 :1)

173 ft lbs of torque on the crank would require 32 psi back pressure on the exhaust brake. The engine would be turning at 3655 rpm and be creating 121 negative HP. The total negative HP required to hold the unit back is the 121 from the exhaust brake plus 50 HP that the wind and drag are responsible for plus the 30 HP absorbed by the driveline.

My truck will maintain 60 mph up a 6% grade. It will not hold back on a 7% downhill grade in 3rd but holds more than enough on a 7% grade in 2nd while pulling a 15,000 lb rv.





Lots of things are missing to do even a crude calculation:


Wind resistance of the rig?

Need square area (meters3) and Cd (Coefficient of drag).


Weight (trailer plus vehicle, both weighted actual, not guess)


Rolling Resistance (actual measured)



The first order factors are different at standstill vs. at speed (say 50mph)

At 0mph, wind resistance is zero or nil

At 50mph it is a major component

Then 2nd order


Then 3rd order things like road surface (which can be 2nd or first order if it is between perfectly flat and smooth surface and 3ft thick mud!)


Fill in the blanks before you do those calculations.




To simplify things --- always calculate net HP measured at the wheel, not gross.

I disagree with direct conversion from fuel used to HP hr as the conversion efficiency varies widely.

Anyone pushing max hp and torque find that the specific fuel consumption (fuel used to generate each hp hr or kwh) rises quite a bit.

* This post was edited 04/10/12 05:58am by NewsW *

4x4ord

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Posted: 04/10/12 05:08am Link  |  Print  |  Notify Moderator

The Mad Norsky wrote:

#40Fan wrote:

Ford EB's aren't very good is the answer to the equation.


I was under that impression also, based almost entirely on that "truck tow out" test where they used I-70 in Colorado and tested Dodge, Ford and Chevy going downhill from the Eisenhower Tunnel.

However, further reading and reviews seem to indicate it is just as good as the other two, albeit with the limiting factor that the starting decent speed on the Ford must be no more than 50 MPH. In the "truck tow test" I believe they started their decent at a speed higher than 50 MPH with all vehicles.


Now back to our regular program. I certainly do not have the know how to contribute any more to this.

I think you and seabiscuit are right in that we need to understand what our respective trucks are designed to do and operate them in that range.

Had Ford or Dodge been conducting the Eisehower pass test they would have chosen a different speed limit. The GM wanted to go down at 60 mph and may have required multiple braking applicatiins to keep it below 50. The Dodge may have been able to maintain 55 mph if it would have been slowed to 45 to get it into 2nd gear and the Ford would likely have been very comfortable at 45 mph in second gear.


2011 F350 SRW short box 4x4 CC 6.7 PS King Ranch
B&W TurnoverBall and Companion
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Gooseneck

Waynesville, NC, USA

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Posted: 04/10/12 05:48am Link  |  Print  |  Notify Moderator

I think we think entirely too much. Does it go? Yes! Does it stop? Yes! Works for me and I have more time to "smell the roses". You'all have a great day, ya heah!


Gooseneck
2003 F350 4DR DRW w/7.3PSD 4WD
Upfitted/Updated Dutchmen 30' 5er


4x4ord

Canada

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Posted: 04/10/12 06:19am Link  |  Print  |  Notify Moderator

NewsW wrote:

4x4ord wrote:

I was playing with the calculator and came up with some figures that seem to make some sense. Based on my experience with the truck I think the numbers are likely right but if there are some numbers guys out there that want to kill a little time I don't mind being corrected.

Based on:
11 mpg towing a 15,000 lb RV at 60mph on a level road with no wind;
and 18 HP hrs/gal (fairly optimistic efficiency for a diesel engine)
6.7 Powerstroke with 3.55 rear axle and 20" wheels
1.52:1 third gear and 2.32:1 2nd gear


These are the numbers I calculated:

HP required to overcome wind drag and rolling resistance and driveline loss is 98 HP on a level road with no wind.

Torque on the crankshaft at 60 mph is 365 ft lbs in 6th gear at 1411 rpm assuming level ground and no wind

lbs of pull between the drive tires and road surface is 490 lbs. (this assumes 20% power loss through the driveline).

Now start climbing a hill. A 7% grade and a 24,000 lb truck and trailer would put an additional 1676 lbs of pull on the surface of the drive tires.
This would equate to 335 hp at the crank (assuming 20% driveline loss)

So the total engine HP required to pull a 15000 lb trailer up a 7% grade at 60 mph would be very close to 433 HP.

To go down at 60 mph there would be the 1186 lbs of pull on the surface of the tires to overcome. This would create 249 ft lbs of torque on the crank assuming 3rd gear and 20% power absorbed into the driveline. To create that torque the exhaust brake would be required to hold 46 psi back pressure at 2920 rpm. The total power required to hold the unit back would be 139 HP from the exhaust brake plus 98 HP absorbed by the driveline, wind etc.

Now if you assume you go down at only 45 mph. The wind resistance would drop drastically so that the power holding the vehicle back due to wind, driveline and rolling resistance might be only 50 hp instead of 98. At 45 mph this equates to 417 lbs of pull on the outside surface of the rear wheels.

The hill and weight of the truck and trailer would still put 1676 lbs of pull at the surface of the pavement and rear wheels. So if you subtract the 417 that the wind and rolling resistance account for, and then account for the power absorbed by the driveline there would need to be 173 ft lbs of torque on the crank in order to hold the unit back from accelerating down the hill. (assuming the truck is shifted down into 2nd gear - 2.32 :1)

173 ft lbs of torque on the crank would require 32 psi back pressure on the exhaust brake. The engine would be turning at 3655 rpm and be creating 121 negative HP. The total negative HP required to hold the unit back is the 121 from the exhaust brake plus 50 HP that the wind and drag are responsible for plus the 30 HP absorbed by the driveline.

My truck will maintain 60 mph up a 6% grade. It will not hold back on a 7% downhill grade in 3rd but holds more than enough on a 7% grade in 2nd while pulling a 15,000 lb rv.





Lots of things are missing to do even a crude calculation:


Wind resistance of the rig?

Need square area (meters3) and Cd (Coefficient of drag).


Weight (trailer plus vehicle, both weighted actual, not guess)


Rolling Resistance (actual measured)



The first order factors are different at standstill vs. at speed (say 50mph)

At 0mph, wind resistance is zero or nil

At 50mph it is a major component

Then 2nd order


Then 3rd order things like road surface (which can be 2nd or first order if it is between perfectly flat and smooth surface and 3ft thick mud!)


Fill in the blanks before you do those calculations.




To simplify things --- always calculate net HP measured at the wheel, not gross.

I disagree with direct conversion from fuel used to HP as the conversion efficiency varies widely.

Anyone pushing max hp and torque find that the specific fuel consumption (fuel used to generate each hp) rises quite a bit.


Maybe its all futile - but I do like to play with numbers.
I have no idea what the rolling resistance and drag coefficient would be which is why I used the fuel consumption as a means of determining those things. I know that engines used to always get their best efficiency ratings (HP hrs/gal) at the rpm where they made maximum torque. When cruising down a level highway with no wind I wouldn't be pushing maximum horsepower but maybe my assumption of 18 HP*hrs per gal is too optimistic - Maybe 16 would be closer? (changing the number to 16 would drop the HP requirement (to overcome drag, rolling resistance and driveline loss) down from 98 to 87. You say the conversion efficiency varies greatly yet I believe all three trucks (GMC, Ford and Dodge) get fairly similar fuel economy while towing. I am wondering if someone actually knows what the conversion efficiency is for these engines while towing down the highway?
The road surface is pavement. The actual weight is very close to 24,000 (23891 lbs)
Frontal surface area would be 102 square feet for the trailer not including the 6 square ft under the trailer. Truck is roughly 34 square feet.

I like the idea of calculating HP at the rear wheels but at some point the driveline power loss would need to be approximated in order to determine the back pressure of the exhaust required to hold the unit back.

Does it seem reasonable that if 98 HP is required to pull the unit at 60 mph that the power required to pull it at 45 would be 50 HP?

It would be interesting if someone actually had a pressure gauge on the exhaust just ahead of the turbo.

NewsW

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Posted: 04/10/12 07:51am Link  |  Print  |  Notify Moderator

Follow method outlined here for rolling resistance estimate:


-------------------------

The Measured Rolling Resistance of Vehicles for Accident Reconstruction

Date Published: 1998-02-23
Paper Number: 980368
DOI: 10.4271/980368

Citation:

Cliff, W. and Bowler, J., "The Measured Rolling Resistance of Vehicles for Accident Reconstruction," SAE Technical Paper 980368, 1998, doi:10.4271/980368.

Author(s):

William E. Cliff - MacInnis Engineering Associates
James J. Bowler - MacInnis Engineering Associates

View All
Abstract:

Knowledge about vehicle rolling resistance is required to calculate speed loss of accident vehicles during portions of their pre-impact and post-impact trajectory when they are not braking or sliding directly sideways. The accuracy of assumed rolling resistance values is most important in accidents with long post-impact roll out distances. Very little hard data are currently available 1 and the accident reconstructionist must usually make estimates of drivetrain losses and normal and damaged tire rolling resistance to determine overall vehicle rolling resistance.

4x4ord

Canada

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Posted: 04/10/12 06:24am Link  |  Print  |  Notify Moderator

Gooseneck wrote:

I think we think entirely too much. Does it go? Yes! Does it stop? Yes! Works for me and I have more time to "smell the roses". You'all have a great day, ya heah!


You could very well be right but then there are others who don't care much for the smell of roses.

NewsW

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Posted: 04/10/12 06:38am Link  |  Print  |  Notify Moderator

4x4ord wrote:



Maybe its all futile - but I do like to play with numbers.
I have no idea what the rolling resistance and drag coefficient would be which is why I used the fuel consumption as a means of determining those things.

Not a sound method because fuel consumption varies widely when measured as specific fuel consumption.



I know that engines used to always get their best efficiency ratings (HP hrs/gal) at the rpm where they made maximum torque.

Not true anymore.

The horsepower race and modern tuning have changed this.

For example, one trick to get more torque out of a diesel is to keep injecting fuel well past TDC even though the fuel have very little time to burn --- but burn a bit it will, and in doing so, sustain cylinder pressure as the piston moves down (more torque) but at a whopping cost to fuel economy and especially specific fuel consumption.

Same tricks can be done with gas.




When cruising down a level highway with no wind I wouldn't be pushing maximum horsepower but maybe my assumption of 18 HP*hrs per gal is too optimistic - Maybe 16 would be closer? (changing the number to 16 would drop the HP requirement (to overcome drag, rolling resistance and driveline loss) down from 98 to 87. You say the conversion efficiency varies greatly yet I believe all three trucks (GMC, Ford and Dodge) get fairly similar fuel economy while towing.

You be surprised. On the modern ones, huge variance depending on how hard it is used.


I am wondering if someone actually knows what the conversion efficiency is for these engines while towing down the highway?

Varies widely.



The road surface is pavement. The actual weight is very close to 24,000 (23891 lbs)

Frontal surface area would be 102 square feet for the trailer not including the 6 square ft under the trailer. Truck is roughly 34 square feet.

Need an actual Cd of the rig.

Measure it experimentally with a wind tunnel.



I like the idea of calculating HP at the rear wheels but at some point the driveline power loss would need to be approximated in order to determine the back pressure of the exhaust required to hold the unit back.

Exhaust back pressure in the relevant ranges will likely be a 2nd or 3rd order effect.


Does it seem reasonable that if 98 HP is required to pull the unit at 60 mph that the power required to pull it at 45 would be 50 HP?

Run the numbers.


It would be interesting if someone actually had a pressure gauge on the exhaust just ahead of the turbo.

It probably is not a big issue.

Say the measured pressure is 2 psi --- what is that as a fraction of cylinder pressure?








http://www.dragtimes.com/horsepower-et-trap-speed-calculator.php


Try this method.

* This post was edited 04/10/12 07:01am by NewsW *

NewsW

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Posted: 04/10/12 07:49am Link  |  Print  |  Notify Moderator

If you want to do an actual measure, do it this way:


http://www.instructables.com/id/Measure-the-drag-coefficient-of-your-car/?ALLSTEPS





Quote:

Measure the drag coefficient of your car
25
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Author:iwilltry(IWillTry.org)
I have a B.A.Sc and M.Eng. from the University of British Columbia, specializing in electromechanical design, but mostly I like to tinker. One of my greatest passions is energy conservation and effici...
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Measure the drag coefficient of your car
i
The purpose of this experiment is to determine your vehicle's drag coefficient Cd and coefficient of rolling resistance Crr. This is done by measuring your vehicle's speed as a function of time while coasting in neutral.

Why would you want to know Cd and Crr for your vehicle? Well, suppose you're interested in modifying your vehicle for improved fuel efficiency. You might consider modifications such as air dams, wheel skirts, removing mirrors, switching to low rolling resistance tires, etc. Cd and Crr offer a quantitative method of comparing vehicle performance before and after these types of modifications to see if you made any improvement.

For other experiments you can do on your car see my website IWillTry.org.
Step 1Equipment
Equipment
i
You will need the following equipment:

a vehicle (and someone with a driver's license)
a clock or stopwatch
a pen and paper (and someone other than the driver to record data)
a flashlight (driving at night avoids traffic)
a long stretch of flat road with little traffic or wind
Excel or another spreadsheet application. I prefer OpenOffice Calc because I like to support open source software, but its Solver function does not handle non-linear systems (yet) so you'll have adjust input variables manually by an iterative process to fit your model to the data (it's not too hard).
The spreadsheet I created to analyze the results. Download here: Drag_Coefficient.xls

Step 2Background Information
Background Information
i
First, let's define some quantities:

Fd is the force on the vehicle due to air resistance (drag) in Newtons
Frr is the force on the vehicle due to rolling resistance in Newtons
F is the total force on the vehicle in Newtons
V is the vehicle's velocity in m/s
a is the vehicle's acceleration in m/s^2
A is vehicle frontal area in m^2
M is vehicle mass including occupants in kg
rho is the density of air which is 1.22 kg/m^3 at sea level
g is the gravitational acceleration constant which is 9.81 m/s^2
Cd is the vehicle's drag coefficient we want to determine
Crr is the vehicle's coefficient of rolling resistance we want to determine

Now for some formulas:

Fd = -Cd*A*0.5*rho*V^2 (formula for force due to air resistance or drag)
Frr = -Crr*M*g (formula for force due to rolling resistance)
F = Fd + Frr (total force is the sum of Fd and Frr)
F = M*a (Newton's second law)

Note that both Fd and Frr are negative indicating that these forces act opposite to the direction of the velocity. Note also that Fd is increases as the square of velocity. This is why driving at high speeds is much less efficient than driving at low speeds. Combining these formulas with a bit of algebra gives us the acceleration due to air and wind resistance as a function of velocity:

a = -(Cd*A*0.5*rho*V^2)/M - Crr*g

Note that the acceleration is negative indicating that air and wind resistance will cause the velocity to decrease.

I created a spreadsheet based on these formulas to generate a model of velocity vs time that can be compared to actual data. The model values for Cd and Crr can thus be adjusted until the model matches the data. This adjustment can be done manually, by overwriting the values of Cd and Crr with new values till the model matches the data, or it can be done using a "Solver" function.
Step 3Procedure
Procedure
i
You can determine Cd and Crr from the same set of test data by measuring velocity with respect to time as your vehicle coasts in neutral. Note that Crr will not be pure rolling resistance but will include some drive-train resistance as well.

1. Drive to a flat road with little traffic or wind.

2. Have the passenger ready with stopwatch and paper to record data.

3. Have the driver accelerate up to above 70 km/h or so, and shift into neutral.

4. Record data as follows. The driver should indicate when the speed drops to exactly 70 km/h. At this time (t=0) the passenger should start the clock. The passenger should indicate every 10 seconds after that and the driver should call out the current speed to the nearest whole km. The passenger should record this value next to each time.

Aside: If you have a digital camera capable of recording several minutes of low resolution video (as most people seem to have these days), the process is much easier and more accurate. You don't need any equipment except the digital camera. Simply have your passenger record a video of your speedometer during the coast down tests, or find some way of mounting the camera so you can do the recording without an assistant. Using a free program such as Avidemux (http://fixounet.free.fr/avidemux/) you can play the video back on your computer frame by frame and view the timestamp at desired speeds.

5. Repeat the test in the opposite direction.

6. Repeat the test in both directions twice more (6 trials in all, 3 in each direction). All these values will be averaged for a more accurate analysis.

7. Download the spreadsheet I created Drag_Coefficient.xls and enter all your data following the instructions included. The spreadsheet averages data from all 6 trials to create a single data set representing velocity (V actual) as a function of time. It then generates it's own model for velocity (V model) based on entered constants and initial guesses for Cd and Crr. Excel's "Solver" function can be used to adjust Cd and Crr in order to minimize the error between the model and actual data. If you are using OpenOffice Calc (which I highly recommend and which you can download for free from http://www.openoffice.org), unfortunately, the solver function currently only handles linear systems, so you will have to adjust the input values manually to minimize the error between the model and the data. Once the error is minimized and the model data matches the actual data as best it can, then Cd and Crr are correct.


Step 4Results
Results
i
Here are the quantities I measured for my car (a 1992 Geo Metro):

M = 1000 kg (about 850kg curb weight plus 150 kg of occupants)
A = 2.3 m^2 (a good approximation based on measurements of my car)

A plot of velocity vs time is shown below. It is based on the averages from my 6 trials. You can see that the model curve closely matches the data points. The values of Cd and Crr for the model are:

Cd = 0.370
Crr = 0.0106

Therefore, these are the drag coefficent and coefficient of rolling resistance for my car.

These values are nice to know. However, in practice, if you want to compare performance before and after making modifications to your car, you can get faster results just by measuring the time to decelerate from speed A to speed B. Pick high to medium speeds if your modifications are likely to affect drag. Pick medium to low speeds if your modifications are likely to affect rolling resistance. Don't forget to take multiple measurements in each direction and average the results.

For more experiments you can do on your car see my website IWillTry.org.

Update 2009-01-02:
I've learned a lot since originally posting this instructable 16 months ago. I've played with measuring Cd and Crr under different conditions on a number of vehicles and other experimenters have picked apart and tweaked my spreadsheet for their own uses.

My experience is that there IS a mistake in one of the underlying assumptions of the model: namely that the force of rolling resistance is constant independent of V. Vehicles are designed with negative lift (so they get pushed into the road more at higher speeds, improving handling) so the force of rolling resistance also has a component that varies with V2 like the drag force. The force of rolling resistance also includes a small component of viscous force (drivetrain) which varies with V.

The model assumes that the drag force is related only to V2 and that the force of rolling and drivetrain resistance is constant. In reality the force of rolling and drivetrain resistance is also related to V2 and V. So a better model of the force on a moving vehicle is:

F = iV2 + jV + k where i, j, and k are constants.

A curve based on that model more closely matches actual coast down data indicating it is a more accurate model. But after solving for i, j and k, there is no way to extract meaningful values of Cd and Crr since by definition, they assume i is related only to drag, and j is 0, neither of which is entirely true.

As mentioned above, If you want to compare performance of a vehicle before and after making mods, the change in coast down time itself is MUCH more meaningful than any change in Cd or Crr extracted from the coast down data.


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