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Discussion Starter #1
Since the amount of torque at the wheels varies with the gear ratio, I was wondering: How much torque is needed to cruise the highway at 65 mph (assume standard sedan shape & flat ground)?


EDIT:
Based upon the Insight tech data posted here, and knowing the Insight barely has enough torque to move itself in 5th gear, the minimum required torque would be at least 85 Nm (2100rpm) * 3.208 final ratio * 0.710 =

~193 Newton-meters to maintain 65mph (142 foot-pounds).

The Civic Hybrid has a similar torque.
Troy
 

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Well, the amount of torque that would be required would be directly related to the losses of the car in question - the main two being aerodynamics and rolling resistance.

"Standard Sedan Shape" is far too vague to give any useful information. Does that refer to a Ford Contour or a Mercury Zephyr? (both having "standard sedan shapes" but very different aerodynamic properties.)
 

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Discussion Starter #3
Think Jetta or Stratus or Civic with 0.29 cd.
 

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It seems ElectricToy is leaning towards Insight performance specifically. I doubt you could port calculations from an Insight to any other "generic" sedan. Not without a hefty penalty to make up for the superior aerodynamics that the Insight provides.

I ran some (extremely) approximate calculations to get a torque value at 65mph. This is doesn't include rolling friction or internal resistances, but I'm guessing those won't add more than 10-20% on top of the force of drag at that speed.

The drag on the Insight at 65mph is about 66 lbs. The "power" applied to the car at that speed by the wind is then 6290 lbs*ft/sec. The torque necessary to equate that power, assuming the engine is turning at about 2000 rpms, would be roughly 30 ft*lbs. Factor in a 10% increase for the all-other-frictions-and-inefficiencies category and you're looking at about 33 ft*lbs.

It's been a while since I've done any of this at school, so let me know if any of these numbers are completely off.
 

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Discussion Starter #5
Rusttree said:
...you're looking at about 33 ft*lbs.
Your calculation is a lot lower than my Insight-engine calculation (142 foot-pounds). So, which one of us made an error?

troy
 
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Hi ElectricTroy:

___The combined output of the Insight’s IMA and the ICE at WOT will not come up to anywhere near 142 Ft-Lb’s. Secondly, the statement that the Insight barely has enough power to move itself in 5th gear is way out there. At 65 mph, lean burn is at the edge of its performance capabilities but the ICE has plenty left beyond that point. Look over some of the top speed threads here and there are a number of members that have seen south of 100 in fifth although I believe they are topping out at 113 mph using third. At 65 mph, the ICE’s torque (ICE only) in whatever gear the torque/HP curve was generated is pushing ~ 85 N*m or 61 Ft.-lb’s at the 2,500 RPM point and no one is into the pack at a 65 mph steady cruise or the IMA would always be dead and a forced charge wouldn’t allow our little beauties to top out at but 50 mph or some other such non-sense if this were the case. Hell, our little beauties are only outputting ~ 23 HP at 2,500 RPM and there is plenty more where that came from.

___Good Luck

___Wayne R. Gerdes
___Hunt Club Farms Landscaping Ltd.
___[email:3ce8traa][email protected][/email:3ce8traa]
 

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I checked out the spec page on the Insight's website (http://automobiles.honda.com/models/specifications_full_specs.asp?ModelName=Insight) and it looks like the max torque (MT) is 79 ft-lbs. That's about half your value of 142. I'm pretty sure the Insight doesn't use twice its rated torque capacity to maintain cruise at 65mph :wink:.

I think the error may be your initial assumption of torque. Your 85Nm converts to about 63 ft-lbs. I think you may have been using the maximum ICE torque as your baseline for your calculation. Is that where you got it from?

--EDIT--
I started writing this post before xcel submitted his post above, so I apologize for the repeated statements.
 

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Discussion Starter #8
xcel said:
___The combined output of the Insight’s IMA and the ICE at WOT will not come up to anywhere near 142 Ft-Lb’s.

Okay here's my math. Where's the error? Thanks Xcel!
85 Nm (2100rpm) * 3.208 final ratio * 0.710 5th gear ratio = _______

troy
 

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As I stated at my last post above, it looks like you're using the max ICE torque (85Nm) as your baseline. The only time you use that much torque is when you're accelerating rapidly, not cruising at highway speed.

Something else that doesn't taste right is multiplying by the gear ratios. I think if you do that, you'll get the torque that the front wheels are applying to the road, not the torque the engine is producing. If you know how much torque the wheels are applying (by doing a force balance or something similar), you can back out the engine torque using:

T2 = T1 * w1 /w2

T1 = torque of the wheels
w1 = angular velocity of the wheels (in rad/sec)
T2 = torque of the engine
w2 = angular velocity of the engine (rpm*2*Pi/60 = rad/sec)
 

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It shouldn't take much more than 20-25hp for most modern cars to hold a steady 65mph on flat ground. I'll use my Civic HX as the baseline. We'll presume it requires 25hp. My Civic turns 2500rpm at 65mph, and since torque = (hp/rpm)*5252 then we can calculate it's making about 52.5 ft-lbs of torque at the crankshaft at that RPM.

If you really, truly wanted the torque at the ground, then you are correct that you'd need to multiply it through the gearing and make a rough estimate for how much is lost due to mechanical inefficiency. Most chassis dynamometers suggest 13% loss for a low power, FWD car, so we'll use that. My HX has a 23.4" diameter wheel, so the radius when accounting for tire squish is perhaps 11.5", or about 0.96 feet. 5th gear is listed as 0.70 and the final drive is 3.72.

The math should thus be 52.5 ft-lbs * 0.70 * 3.72 * 0.96 = 131 ft-lbs. If you figure in 13% loss, then only about 114 ft-lbs actually made it to the ground.
 

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Discussion Starter #11
AZCivic said:
The math should thus be 52.5 ft-lbs * 0.70 * 3.72 * 0.96 = 131 ft-lbs.
That comes close to my 149 ft-lbs for the Insight.

0.70 = 5th gear ratio
3.72 = final drive ratio
Where did that number 0.96 come from?
 

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Discussion Starter #12
Rusttree said:
you're using the max ICE torque (85Nm) as your baseline. The only time you use that much torque is when you're accelerating rapidly, not cruising at highway speed.
The torque curve says 85 @ 2100 rpm. I don't understand why that's wrong?

I thought my insight IS maxed out during highway speed. It can't even climb modest slopes on engine-only.

troy
 

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You are neglecting that the accelerator is not floored to maintain highway speeds.. Use of the IMA is not an indication that the ICE is "maxed out"
 

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Dear Troy

What you seem to have calculated is the torque at the wheels neglecting friction at WOT. There is nothing wrong with your math if that is what you intended to do. Your original question cannot be answered by anyone until more information is givin or assumed. When you ask for torque you must indicate where you are taking the measurement. Torque is not the same at every point in a drive train. In fact the entire point of the drive train is to increase the torque at the driven wheels. The engine on all cars I've ever heard of (including cars with overdrive) turns more often than the wheels which increases torque.

Gear Ratios:
1st: 3.461
2nd: 1.750
3rd: 1.096
4th: 0.857
5th: 0.710

Final Drive Ratio: 3.21

Overall gear ratio in 5th gear is 3.21 times .710.

Not to be overly rude but clearly you understand very little of all this and you are just plugging numbers into a formula and wondering why the answer is not what you expected.

The assumption that the car cannot go over 65 mph is completely baseless as has already been mentioned.

Where did that number 0.96 come from?
so the radius when accounting for tire squish is perhaps 11.5", or about 0.96 feet
 

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ElectricTroy said:
The Civic Hybrid has a similar torque.
Troy
No it does not!

The Civic has an extra cylinder over the Insight.

Civic: Torque: 116 ft-lbs. max at 1500 rpm

Insight: Torque: 79 ft-lbs. max at 1500 rpm
 

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Looking at this from another angle, the Insight can travel at 113 MPH, and since this is limited electronically it does not represent the maximum torque available. Since the Insight can achieve this at red line in third gear We could use the torque figures for red line. Since the air resistance goes up as a squared ratio, the torque required at 60 MPH would be roughly 4 times smaller. Thus the torque would be less than 20 newton - meter, probably significantly less, and less than 19 Horsepower. In theory if the IMA was big enough a single cylinder 250 CC engine would be overkill. :D Scary!
 

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Looking at this from another angle, the Insight can travel at 113 MPH, and since this is limited electronically it does not represent the maximum torque available. Since the Insight can achieve this at red line in third gear We could use the torque figures for red line. Since the air resistance goes up as a squared ratio, the torque required at 60 MPH would be roughly 4 times smaller. Thus the torque would be less than 20 newton - meter, probably significantly less, and less than 19 Horsepower.
Best guess I've seen so far. Not to be nit picky but the torque figure you are refering to is at the engine. Troy looked as if he was attempting to figure the torque at the wheels or he would not have included the final drive ratio.
 

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Thanks Lakedude, I'm not sure what theoriginal question was leading to. If it was to find the required torque at the wheel for an application of electric wheel motors then the torque would have to be divided by two. The same would be true if Electric Troy was trying to determine the stress on the axles. I'm a little curious myself to know how much power it takes to move the Insight at 60 MPH.
 

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Ok my turn

From Edmunds:

Insight

Width: 66.7 in.
Height: 53.3 in
With 4 inches of gound clearance.

Or about 20 sq feet. Guestimating on the "roundness"

Drag = 1/391 CdAV² V in mph, and drag in pounds of force...

Cd is .25

So 65*65*20*.25/391= about 58 pounds drag.

The tires 11.22 inches or .935 feet so the torque at the wheels is about 62 foot pounds.

62/(3.21*.710) = about 27 foot pounds at the engine neglecting driveline losses and other sources of drag.
 

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Discussion Starter #20
I'm going with "about 150 newton-meters at the wheel" to maintain 65mph.

Thanks,
troy
 
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