Assuming you are in 5th gear, the engine is produceing app. 20 ft. pounds of torque and probably 15BHP. 8)

HTH
Willie

This is based on the HP torque graph in the "Technicians Information Guide" booklet I have.

 

I did a test on a windless summer day. I chose a very long straight hill with if memory serves a 3 degree slope and zero traffic, started at the top at about 55 MPH and was doing 70+ and still accellerating at the bottom. In any case I calculated from this rather crude test that 50 MPH should be attainable with less than 12 HP. Testing for 30 MPH should be much safer if you can just find a very long hill with perhaps a one degree slope, no wind and no traffic. I posted the actual test results on one of the threads. You need a good transit or GPS and a little spare time. For accuracy you need to find a speed you can start down the hill and maintain in neutral with just the slope of the hill. Calculate the force due to gravity using a log table or scientific calculator. Multiply force X distance X time, then convert to HP using an Internet online converter. The toughest part is finding the right hill.

 

Interesting I thought 15bhp was bit high, my 2200lb ev with the aerodynamics of a brick and a 12kw motor could do 50mph on the flat. I would suggest 3-4kw might be nearer the mark.

 

There was a corner at the top of the hill that I wasn't comfortable taking at more than 55. The hill at 3 degrees was just too steep for a real test without feeling like I was going for the sound barrier, :roll: and anything other than a steady speed versus constant slope requires too much rocket science (calculus).

 

How much power is reqd to maintain an Insight at 30mph on the flat in good calm conditions? BHP? KW? Assume tyres at say 50psi. Any offers?

30 mph = ~48.28 kph = ~13.411 m/s

The formula for Instantaneous Rolling resistance force in ( Newtons ) N is:
F = -Cr * m * g (rolling resistance)
F = Force of rolling resistance ( Newton )
Cr = Coefficient of rolling resistance of LRR Tires on a smooth road 0.006-0.01
m * g = Total Weight
g =9.8 m / s^2
m = 834 kg for a 5 speed Insight without AC.

So under good conditions the Instantaneous Rolling resistance = 834*9.8*0.006 = ~49.0392 N
1 Watt = 1 Nm/s
49.0392 N * 13.411 m/s = ~657.6647112 Watts

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

Aerodynamic Drag :
Instantaneous force of drag is:
P = -1/2 * p * V * V * A * Cd
where :
F is the force of drag ( Newton )
p is the density of the Air ( kg/m³ )
V is the velocity of the car ( meters per unit time )
A is the frontal Area ( Insight = 1.9 Square Meters )
Cd is the drag coefficient ( Insight = 0.25 )
The air density will change with humidity and temperature as will Cr, but at about...
At +30 degrees C the Density of the air is about 1.164 kg/m³

Instantaneous Aerodynamic drag force:
1/2*1.164*13.411*13.411*1.9*0.25 = ~49.721 N
1 Watt = 1 Nm/s
49.721 * 13.411 = ~666.807 Watts

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

~657.6647112 Watts Rolling resistance + ~666.807 Watts Aerodynamic Resistance = ~1,324.47161 Watts
= ~ 1.3245 kW
1kW = ~1.34 HP

So it takes a minimum of ~0.9884 HP ( just about 1 HP ) for an insight to maintain about ~30 MPH on a flat level ground under good conditions and ~30 degrees C.

Of course real world variable will cause variation.

 

Ian is right that it takes at LEAST 1 HP to maintain 30 mph, but his calculations are missing other important factors such as mechanical frictional losses etc..

A better way to determine the power required to drive at a specific speed is to do a coast down test on a level road with no wind.
Since finding a completely level road and no wind (therefore no traffic as well) is extremely difficult, just find an empty stretch of road that comes close to that description and do coast down testing in both directions multiple times to average out the error.

One of Newtons laws says that an object in motion will remain in motion until an external force is acted on it. (something like that)
What this means is that there are forces acted on the Insight that prevent it from maintaining a constant speed on level ground unless it is outputting some power to overcome air resistance, rolling resistance and mechanical friction in the bearings etc...

The main equation to use is
Force = Mass x Acceleration

Mass can be measured at truck weigh station or at a dump weigh station or just use the published car weight plus driver and passenger/cargo weight.
Acceleration can be calculated by plotting the data from the coast down test.

Coast down test:
Use a passenger to help to improve accuracy and for safety reasons
Use a stopwatch with slip times or lap time memory
Use a level straight stretch of road with little or no wind and no traffic (because you will be slowing down a lot)
Start at a high speed for example 60 mph.
put the car in neutral and have the passenger stare at the speedometer.
Have the passenger start the stopwatch as the car speed passes exactly 50 mph and measure "lap time" at every 10 mph interval.

Acceleration = Change in Velocity / Time

Rolling resistance and mechanical friction changes very little with speed.
But wind resistance increases exponentially compared to the increase in speed.
Therefore at higher speeds such as highway driving wind resistance is significantly more then rolling resistance/friction.
But at low speeds wind resistance can be less significant then rolling resistance/friction.
So during the coast down test the time measured increases between the 10 mph drop intervals.
So the calculated Acceleration will the be the average acceleration between the speed interval range.
Since this is a coast down test when the vehicle is in neutral, the calculated power is the engine power applied to the clutch to maintain a constant speed.
So the calculated power is closer to crank HP then wheel HP.

I haven't done this before so I don't have numbers to work with, but hopefully I explained in a way that anyone that remembers high school level physics can continue where i left off.

Let me know what you find out.

 

I dont know the accuracy, but the ScanGauge and other scan tools will give you a real-time number for Horsepower at any given instant. I assume it is ICE horsepower only, so run your tests without Regen or Assist.

My 4,000 pound shaped like a brick with huge tires Escape Hybrid with eCVT takes the following:

22 HP to maintain 55 MPH
33 HP to maintain 65 MPH
44 HP to maintain 75 MPH

50 HP will maintain 55 MPH up a 6-7% grade.

I've only had a scangauge in the Insight for a few days and have not run the same tests yet.
If there are interested parties, I would be happy to repeat the above.
Often, I see numbers in the single digits when drving the Insight.
8) -John

 

Ian is right that it takes at LEAST 1 HP to maintain 30 mph, but his calculations are missing other important factors such as mechanical frictional losses etc..

100% correct... the 1 HP is the amount of power needed to be applied to over come the aerodynamic and rolling resistance under those limited conditions.... the Transmission is not 100% efficient so losses.... etc... this means you have to generate more than 1 HP to get 1 HP worth of applied power.....

Also going up even a 1% incline will change the results as well...
rolling resistance will be reduced by 1% as the normal force against the road is no longer the whole weight...
so ~6.57 Watts less rolling resistance... but you traded it for gravitational force acting against you... which means ~1% of the gravitation force of the car becomes a counter force... or about 834*9.8*0.01= ~81.732 N * 13.411 = ~1,096 Watts = ~1.1kW of gravitational resistance to even a 1% incline.

and it assumes a wind speed of zero... faster winds speeds will increase the aerodynamic resistance exponentially.

Like I said in the calculation post ... "real world variables will cause variation."

Unfortunately it is nearly impossible to achieve the limited "Good" conditions in the real world... Wind speed will only be near zero in a large warehouse with all the doors closed... you would also need such a wear house with a perfectly level floor as even small variations in angle will throw off the results.... etc...etc... the real world is tooo complicated. :?

 

Not a very exact test, but should give you some guidelines.
My e-wheel has taken me 40+ miles at 24-30MPH on a full charge 100% to 40% SOC of my AGM31 110AH 48V pack. The motor current is limited to 190A. Cruising at 30 mph, flat ground, it sits at about 20-30A. The road I traveled during this test run included the stretch of road displayed in this topo graph, http://www.99mpg.com/Projectcars/TheBes ... file.jpg,0

I would guess that no more than 2-3HP would be required. The acceleration with that small amount of power will be painfully slow though.

 

For another data point, I calculated approximately 17 hp to maintain 60-65 mph. Did it the same way, with a coast down a long slope - about 10 miles and 2000 ft elevation change. The actual figure is probably a little less, as the road curves and so there are extra friction losses, plus a bit of regen etc.

Also, I'm curious as to how to get a horsepower figure out of a ScanGauge...

 

So a prius 40 HP brushless motor should get the car up to 80??

 

Mike, 40 HP should get you going faster than 80, perhaps a little over 90 if you have long enough to wait. I found the test I did:

viewtopic.php?f=1&t=4555&hilit=+slope

I had asked the question: Horsepower to the wheels at 60 MPH?

Here is my conclusion, (although I believe the power will prove to be less than my test indicates as I believe I assumed that all losses were due to wind resistance):

1 horsepower = 550 foot-pounds per second

76 MPH = 111.5 feet per second.

(Sine 3 degrees) .0523359 x 111.5 feet per second = 5.83 feet per second vertically.

Insight and pilot weigh 2,049 pounds x 5.83 = 11,946 vertical foot pounds per second

Divide by 550 = 21.7 horsepower at 76 miles per hour.

Finding the cube root of 21.7 (2.79) and dividing by the ratio of various speeds, then cubing the result yields:

(at 50 degrees)

6.2 horsepower at 50 MPH
10.7 horsepower at 60 MPH
25.3 horsepower at 80 MPH
49.5 horsepower at 100 MPH
71.3 horsepower at 113 MPH

 

So a prius 40 HP brushless motor should get the car up to 80??

I'm with b1shmu63 (kip) ... although my math bellow looks a bit different... and the numbers I got are a bit different from his... I still think 90+ MPH is real world possibility from 40HP in an Insight.

If you have the power to drive it at full power, I would guess around 100mph in real world driving conditions... of course many variables will change the results.

:!: Beware of Math :!:

40 HP = ~53.6 kW = ~53,600 Watts of produced power - Transmission and other losses to apply power. Most transmissions are very efficient probably less than 5% loss. Rolling resistance , aerodynamic drag , and gravitational counter force are the biggest three forces as far as I can see.

~53,600 - 5% Transmission and other minor losses = ~50,920 watts applied or ~38HP

Of course to see 53.6 kW from the motor even at 90% efficient motor will pull ~59.56kW from the Batteries... Which is allot of electrical energy.... even at ~200Amps that is ~300V.

With 0 wind...+30 C ... V >> 0% incline >> 1% incline >> 2% incline
10 m/s = ~22.37 MPH >> ~766.84W (~0.57HP) >> ~1,579.26W (~1.18HP) >> ~2,391.67W (~1.78HP)
20 m/s = ~44.74 MPH >> ~3,192.38W (~2.38HP) >> ~4,817.22W (~3.59HP) >> ~6442.05W (~4.81HP)
30 m/s = ~67.11 MPH >> ~8,935.33W (~6.67HP) >> ~11,372.57W (~8.49HP) >> ~13,809.82W (~10.31HP)
40 m/s = ~89.48 MPH >> ~19,654.37W (~14.67HP) >> ~22,904.03W (~17.09HP) >> ~26,153.7W (~19.52HP)
50 m/s = ~111.85 MPH >> ~37,008.21W (~27.62HP) >> ~41,070.29W (~30.65HP) >> ~45,132.37W (~33.68HP)

Of course real world variables will cause variation.... but I would think this to be within +/- 2% of real world under the given conditions.

The above comes from the bellow:

The formula for Instantaneous Rolling resistance force in ( Newtons ) N is:
F = -Cr * m * g (rolling resistance)
Instantaneous force of Aerodynamic drag in N is:
P = -1/2 * p * V * V * A * Cd

Together they form ( Total resistance [ in N ] ) Tr = ( Cr*m*g ) + ( 1/2*p*V*V*A*Cd )
Cr = Coefficient of rolling resistance of LRR Tires on a smooth road 0.006-0.01
g =9.8 m / s^2
m = 834 kg for a 5 speed Insight without AC.
A = the frontal Area ( Insight = 1.9 Square Meters )
Cd = the drag coefficient ( Insight = 0.25 )
p = The Density of the air. At +30 degrees C it is about ~1.164 kg/m³

Tr = 0.006 * 834 * 9.8 + 1/2 * 1.164 * V * V * 1.9 * 0.25
Tr = 49.0392 + 0.27645 * V * V

1kW = ~1.34 HP

Wind:
only effects Aerodynamic drag.
Head wind adds to V .... V = 10 m/s + 10m/s Wind >> Va = 20 m/s
Tail wind subtracts from V .... V = 10 m/s - 2m/s wind >> Va = 8m/s

Gravity:
as you go up an incline your normal force for the rolling resistance will be decreased.
So a 1% incline will reduce your rolling resistance to 99% of the 0% incline rolling resistance.
while at the same time adding 1% of the vehicle weight as a pull against you... 834 * 9.8 = 8,173.2 * 0.01 = 81.732 N

1 Watt = 1 Nm/s

Total power ( Tp ) of rolling resistance , aerodynamic drag , and gravitational pull at given conditions:
Tp = [ ( 49.0392 * V )*( %incline @ 1% = 0.99 ) ] + [ 0.27645 * Va * Va * Va ] + [ V * 8,173.2 * ( %incline @ 1% = 0.01 ) ]

 

"I still think 90+ MPH is real world possibility from 40HP in an Insight."

I have to disagree, at least in practical driving. Sure, if you take it out to Black Rock (http://en.wikipedia.org/wiki/Black_Rock_Desert ) or some such place where you have 40 miles or so of absolute flat surface - but then, landsailers have been known to do 100+ mph with no engine at all

Unfortunately for fuel economy, the world is not entirely covered with dry lake beds, and so gravity comes into play. 1% grades are not the usual thing in this part of the world. 5-7% is common on main highways, smaller roads will have steeper. And of course there are places in Europe where you'll find 18-20% or more. (I swear some of the roads in Yorkshire & the Lake District are over 50%, but being on a bike may have distorted my perspective a bit.)

This really gets back to the whole hybrid rationalle. You don't need much power to go down a level road at constant speed. You need it to accelerate, and to climb hills.

 

Correct James.

It would likely take several miles to approach that speed. The extra mass of the batteries would not help either for acceleration. I note that the GM engineers are extending the front of their Volt plug in series hybrid design.

http://www.greencarcongress.com/2008/04 ... .html#more

The same could be done on the Insight to improve airflow around the body. Underneath the Insight there would be no Catalytic converters and muffler so it should be possible to smooth the air flow there as well. In any case it is not usually necessary to exceed 60 MPH by much. The electric motor would have terrific torque from a standing stop so it would feel responsive around town. The batteries....well yes, that is a recurrent issue.

 

"I still think 90+ MPH is real world possibility from 40HP in an Insight."

I have to disagree, at least in practical driving. Sure, if you take it out to Black Rock (http://en.wikipedia.org/wiki/Black_Rock_Desert ) or some such place where you have 40 miles or so of absolute flat surface - but then, landsailers have been known to do 100+ mph with no engine at all

Unfortunately for fuel economy, the world is not entirely covered with dry lake beds, and so gravity comes into play. 1% grades are not the usual thing in this part of the world. 5-7% is common on main highways, smaller roads will have steeper. And of course there are places in Europe where you'll find 18-20% or more. (I swear some of the roads in Yorkshire & the Lake District are over 50%, but being on a bike may have distorted my perspective a bit.)

This really gets back to the whole hybrid rationalle. You don't need much power to go down a level road at constant speed. You need it to accelerate, and to climb hills.

I agree with you that there is always a place with a steep enough hill that 40HP will not get you to 90+ MPH... but since the given conditions are a flat and level 0% grade I will also still stand behind my claim of 90+MPH from a Insight Which I think it should be able to do at over 2% grades.

as for 5% grades...

:!: Beware of Math :!:

Tp = [ ( 49.0392 * V )*( %incline @ 1% = 0.99 ) ] + [ 0.27645 * Va * Va * Va ] + [ V * 8,173.2 * ( %incline @ 1% = 0.01 ) ]

40 HP should give about ~50,920 watts applied or ~38HP to move the car after transmission and such losses.

with 0 wind Va = V.... at +30C

50,920 = ( 49.0392 * 0.95 * V ) + ( 0.27645 * V * V * V ) + ( 8,173.2 * V * 0.05 )

Solve for V = just over ~47.35 m/s or ~105.9 MPH ( I figure +/- 2% ) ... so... I still say even at 5% incline I would expect a 40 HP EV Insight to do 90+ MPH even in real world conditions.

But we will only know when it is done ... if you can get the ~60kW or so of electrical power from batteries to feed it to full power.

 

How much power is reqd to maintain an Insight at 30mph on the flat in good calm conditions?
Assume tyres at say 50psi.
Any offers?

And the answer is....... about 6.5 HP.

I just went out with 50psi tires on the flattest road I could find. Keep in mind this was at 4700ft elevation.
This was electronically measured with a ScanGauge. One 200 lb driver, no cargo.

The car makes 2.8 HP at Idle.

30 MPH and 1050 RPM = 6.5 HP
40 MPH and 1370 RPM = 9 HP
50 MPH and 1700 RPM = 12 HP
60 MPH and 2080 RPM = 15.5 HP
70 MPH and 2400 RPM = 20 HP

-Enjoy!
-John

 

Divide by 550 = 21.7 horsepower at 76 miles per hour.

Finding the cube root of 21.7 (2.79) and dividing by the ratio of various speeds, then cubing the result yields:

6.2 horsepower at 50 MPH
10.7 horsepower at 60 MPH
25.3 horsepower at 80 MPH
49.5 horsepower at 100 MPH
71.3 horsepower at 113 MPH

Kip, your theoretical value for 76 / 80 MPH is very, very close to reality! Good Job!
It takes a little more than you projected for 50 and 60, though.
I was getting consistent measurements of 11 to 13 horsepower following gentle contours of the road at 50 MPH.
And that was with thinner air @ 4700ft elevation. ( Colorado ) -John

 

yep, 13 to 14 at around 55 mph, per SG2. 35 is a speed i rarely cruise at, so i dont have a feel for that one.

 

2.8 HP at idle is particularly interesting as it illustrates the difference between neutral coasting with the engine idling and engine off idling. In my calculations for lower speeds I had assumed all losses to be wind resistance which was certainly erroneous, thus my optimistic results. This data is extremely valuable for those planning electric conversions as well as those calculating the gains from energy saving mods and driving techniques. Thanks for everyone’s input and for any additional data we can post here. We might finally be able to answer the question as to whether it is better to downshift to fourth gear when cruising at 30 to 40 MPH due to transmission losses for in overdrive and efficiency losses due to running the ICE at less than optimal RPM.

 

How does the ScanGauge calculate engine power?
Does it require the driver to do a coast down test?

 

I'm not sure how it does it internally.
There is no calibration other than you need to have the main setting to 1.0 liter engine.

You program this xgauge on xgauge capable units:

TXD 00
RXF 400080000000
RXD 0000
MTH 000A00170000
NAM HPR

 

If I could find a long flat road, (not easy in NE) I should be able to just run with the e-wheel @ 30MPH and measure the voltage and current delivered to the motor to do a real world HP measurement. I can run the motor alone with the wheel lifted to get a baseline current.

The losses in a toothed belt 4:1 reduction should be quite small, and the motor torque and efficiency curves of the E-Tec motor are here: http://www.thunderstruck-ev.com/etek_curve.htm

 

I wonder if the scangauge calculates HP using instantaneous fuel consumption, air mass flow into the engine and temperature along with some other info.
I imagine this method might provide a HP value maybe within 20% or 30% of reality which is not bad.

 

I'm sure it does something like that and I have used this in 4 different models of car, and I would say it is very close to actual. The HP does not go down during lean burn, so only trust the numbers under regular burn conditions. -John