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Old 01-18-2023, 05:34 AM   #1
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How much Horse Power do YOU need?

How much Horse Power do you need? The answer requires a consideration of the resistance category and the related factors.

For resistance, the categories are; Air Resistance; Rolling Resistance; Acceleration; Slope.

For factors, the components are: coefficients of resistance and air density; surface area; mass; velocity; time.

Attached is a pdf for very basic assumptions and calculations, in perfect conditions.
Assumed is a rig having GCWR of 9000lbs, or 4000KG.

DRAG power assumes no head- or cross- winds.
* Mass is not a factor for Drag.
* Air Density is the value at seal level.
* The Drag Coeficient is estimated at 1.0. Passenger cars have a coef of about 0.3. Trucks are about 0.6. Nobody has published a coef for the average trailer. The coef depends upon the aerodynamic shape of the vehicle. Most trailers are little more than a box, which has a minimum coef of 1.0.
* Drag is affected signficantly by frontal area. A TV with a TT will have at minimum a frontal area of 5m2.

(Note. Some TT brands, such as Airstream, make big claims their products are areodynamic. Such claims are highly dubious, and never supported by wind tunnel trials. The primary factor for reducing drag while in motion is to ensure a laminar flow. No truck can produce a laminar flow: trucks produce enormous turbulence, which flows rearward to the TT. The turbulence increases drag greatly, which no TT can magically transform into a laminar flow. Thus, no matter what the shape of the TT, turbulence is the biggest factor.)

ROLLING RESISTANCE power depends upon the rolling resistance of tires, suspension, transmission, and mass.
* Passenger cars have the least RR, at about coef 0.01. Trucks have about 2x more RR, say 0.02.

ACCELERATON power depends upon the desired Time interval. Assumed are: acceleration from 0 to desired velocity (eg 0 to 50 mph); and the Time is pegged at 20 seconds.

SLOPE power depends upon the grade. In the US, the average grade is 6%, so the slope is 0.6.

So what do the numbers mean? Depends upon the chosen velocity.

If the rig is 4000kg, and has a frontal area of 5m2 ...
* the HP needed to overcome Drag and RR to maintain 50mph, is 89HP.
* the HP needed to overcome Drag and RR to maintain 60mph, is 106HP.
* the HP needed to overcome Drag and RR to maintain 65mph, is 1306HP.

Thus, 65mph requires 46% more HP than 50mph, in perfect conditions of no wind.
When Drag + RR + Slope are tallied, the power to maintain 65mph up a slope is more than 2x the power needed for 50mph.

How this relates to fuel consumption is a very complicated relationship, for nothing is linear. How each engine produces say 200HP will depend upon many non-linear factors. BUT if the same engine is used, the differences in HP noted above do translate into differences in fuel consumption. In the most simplest terms, a velocity that requires 2x the power will require at least 2x the fuel consumption.

People may wish to consider their chosen speeds when commenting upon the increased costs of RV-ing. Speed is one factor that can translate into less fuel consumption, and thus less cost - all factors considered.
Attached Files
File Type: pdf HorsePowerRequirements.pdf (25.8 KB, 12 views)
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Old 01-18-2023, 07:24 AM   #2
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If your assumptions are correct it would take about 250 HP to tug a typical 16000 lb 5ver down the highway at 60 mph. I think it takes about half that amount of power.
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Old 01-18-2023, 08:18 AM   #3
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Horsepower is not the total answer. The engine need torque to pull a load.

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Old 01-18-2023, 08:23 AM   #4
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I don’t know how much I need, but I know how much I like. . Our SUV has about 450 hp and around the same torque. It pulls our 3400 pound loaded trailer effortlessly. But really it’s overkill. Our previous SUV was probably half that and it did alright although lots of revving and gear changin on the hills.
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Old 01-18-2023, 09:52 AM   #5
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new diesels

I would like HP but torque is what you need to haul these big 5th wheels. My 2006 Dodge Ram 3500 5.9 Cummins diesel had 325 HP and 610 LBFT of torque. It pulled my 13500 lb 5th wheel good but struggled on long grades. Without a exhaust brake it was a chore on long down hill grades. It did ok passing but took it time building power. I moved to a 2017 F350 DRW diesel with 440 HP 925 LBFT of torque. Climbing hills is effortless and the exhaust brake work great. I have 4.10 gears so that might help with the exhaust brake. I can set the cruise control at 70 MPH and the exhaust brake in auto and it is a comfortable tow. I can maintain speed on any hill and the truck is not struggling. Now my truck is at 54% GCWR. I have more truck then what I need but it handles the camper like it is on rails. My gauges shows running down the highway I am at 400 to 500 LBFT of torque. On a hill I am at 900 + LBFt of torque.
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Old 01-18-2023, 10:00 AM   #6
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Interesting post (pst...fix your HP at 65mph; you have an extra digit there). While it's fun to read, the actual results are only as good as the info and assumptions inputted and the more assumptions that are made, the farther the guess is from reality (propagation of errors).

Still fun and illustrates the effect speed has on the power requirements. All I know is that my 475hp, 1050 ft lb engine seems to have no problem maintaining 60 - 65mph with my 18,000 GVW on local highways and passes.

2022 Outdoors RV 25RDS, 2022 F350 dually, 6.7PSD, 10 spd, 3.55's
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Old 01-18-2023, 10:57 AM   #7
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The title of the tread is: How much Horse Power do YOU need?

Most of the answeris more about YOUR EXPECTATIONS than any actual mechanical "need".

We have had DP with 250 HP, 300 HP and 400 HP with no significant difference in weight. Every one of them got us exactly where we wanted to go.

If you have to have enough HP so that you climb grades with the cars, the answer is very different than if you are satisfied with just being faster than the loaded 18 wheelers.

So, your expectations are the key.
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Ex: 2003 Alpine 38FDDS. Ex: 1997 Safari Sahara. Ex: 1993 Foretravel U240
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Old 01-18-2023, 11:28 AM   #8
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I agree you don't need much HP. I have had cars with 90 HP and they drive fine. Then 116 HP car for years. These cars did not have much torque either.

I did have a V8 Mercury Cougar XR7 that had 150 HP and 270 Ft lbs of torque. That was a good driving car with about the right amount of power.

So I agree you do not need much HP.
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Old 01-18-2023, 01:03 PM   #9
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Don't give a damn about horsepower. Show me the torque curve. That's what moves the load. I want to know how much torque an engine makes between idle and 2000 rpm where the engine spends most of its time.

Why would I care if an engine makes 400 HP and 450 lb-ft of torque at 5000 rpm? Meaningless, 'cause my engines aren't going to be running at that rpm.
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Old 01-18-2023, 02:02 PM   #10
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^^^The peak torque and Hp figures are usually stated at a particular rpm. When considered together they give a pretty good idea of the engine’s power curve.
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Old 01-18-2023, 02:58 PM   #11
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You need to worry about HP or else you end up with a bulldozer. HP should be about half the torque number. 500HP with 1,000 ft lbs of torque would work for me. But HP could be 400 and still work nice.
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Old 01-18-2023, 03:51 PM   #12
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My 2007 Dodge with 5.9L CTD with 325HP/2900RPM and 610LBFT/1600RPM 4 speed tranny with 4:10 differential
Pulls my 13,838# tall 5th wheel great and haven't found a grade that it struggled the 4" Exhaust Brake handles downgrades with ease

I'm good with its HP/Torque --------
I took my Medication today. HAVE YOU?
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Old 01-18-2023, 04:41 PM   #13
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Assuming a drag coefficient of 1 gives much too high of value so I think using a drag coefficient of .6 for a typical fifth wheel will give a more accurate result. And maybe the rolling resistance for the truck and rv tires should be more like .015. The other thing is that the calculated values are rear wheel hp. The manufacturers rate their engines at the crankshaft.

If it really took 235 HP to tow a 25,000 lb truck/RV down the highway we would only be getting 5.5 mpg.
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Old 01-19-2023, 05:41 AM   #14
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I did a little number crunching this morning and came up with what I think might be a more useful horsepower formula. (This should provide a fairly accurate number especially for a diesel powered truck)

Horsepower required to overcome rolling resistance and drag = 21.3 x speed /fuel economy at that speed on level ground.

Crankshaft horsepower required to climb a grade = (grade x speed x weight of truck and trailer /31875) + hp required to overcome rolling resistance and drag

So to climb a 7% grade at 45 mph with a 25000 lb truck and trailer would take:

(7 x 45 x 25000)/31875+ (21.3 x 45/11) =

247 + 87 = 334 hp
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