Would WHP/tonne be more accurate when comparing acceleration??
23 June 2006, 02:00 PM
#1
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Would WHP/tonne be more accurate when comparing acceleration??
We usually look at BHP per tonne when we try to gauge a car's acceleration. of course there are also factors like drag, traction, gearing etc etc...but what about the actual WHP to weight instead of BHP to weight?? Wouldn't it be an even more accurate gauge??
for eg.
civic with 180whp
weight 1100kg
WHP = 163.63 whp/tonne
VS
wrx with 200whp
weight 1400kg
WHP = 142.85 whp/tonne
now, the civic would probably have around 220-225 bhp say 15% tranny lost on a FWD, whereas the wrx would need around 265-270 bhp to achieve 200 whp with 23-25% tranny lost...would the civic be faster in both drag and in gear acceleration??
for eg.
civic with 180whp
weight 1100kg
WHP = 163.63 whp/tonne
VS
wrx with 200whp
weight 1400kg
WHP = 142.85 whp/tonne
now, the civic would probably have around 220-225 bhp say 15% tranny lost on a FWD, whereas the wrx would need around 265-270 bhp to achieve 200 whp with 23-25% tranny lost...would the civic be faster in both drag and in gear acceleration??
23 June 2006, 02:21 PM
#5
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Thread Starter
i dynoed my jdm civic sir on the dynojet and got around 143-144 whp which is around 15% down from the factory 170hp rated at the crank.
why wouldn't this make sense??
4WD does take away a whole load of HP...
why wouldn't this make sense??
4WD does take away a whole load of HP...
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23 June 2006, 02:23 PM
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Originally Posted by Suberman
i dynoed my jdm civic sir on the dynojet and got around 143-144 whp which is around 15% down from the factory 170hp rated at the crank.
why wouldn't this make sense??
4WD does take away a whole load of HP...
why wouldn't this make sense??
4WD does take away a whole load of HP...
23 June 2006, 02:29 PM
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This should help ! DUR ! thicko !
For example the velocity (v1) at one second (t1) is;
v1=((a0+a1)/2 *Delta t)+a0=((0.00+12.868)/2) *1)+0 = 6.434
v2=((a1+a2)/2 *Delta t)+a1=((12.868+25.736)/2)*1)+6.434=
25.736
etc.
If we plot the velocity as a function of time we obtain
the distance (d)traveled.
For example the distance d1 at t1 is
d1=((v0+v1)/2 *Delta t)+d0=((0.00+6.434)/2)*1)+0 = 3.22
d2=((v1+v2)/2 *Delta t)+d1=((6.434+25.736)/2)*1)+3.22=
19.30
etc.
Time__acceleration____velocity-MPH_distance
(sec__Gs_(ft/sec^2)___(ft/sec)_MPH_____(ft)
t0___0.0____0.00____0.000___0.00____0.00
t1___0.4___12.87____6.434___4.39____3.22
t2___0.8___25.74___25.736__17.55___19.30
t3___1.0___32.17___54.689__37.29___59.51
t4___1.0___32.17___86.859__59.22__130.29
t5___0.7___22.52__114.204__77.87__230.82
t6___0.6___19.30__135.114__92.12__355.48
t7___0.4___12.87__151.199_103.09__498.64
t8___0.2____6.43__160.850_109.67__654.66
t9___0.1____3.22__165.676_112.96__817.92
t10__0.0____0.00__167.284_114.06__984.40
t11__0.0____0.00__167.284_114.06_1151.69
Step 2. Using the velocity and fixed mass (m) of the
car, you then calculate the kinetic energy of the car
as a function of time.
Kinetic energy (KE)equals the car’s mass (m)divided by
2, times the velocity (v)squared (^2))
KE = (m/2)*v^2
t0 KE0 = 0
t1 KE1 =(93.24/2)* 6.434^2 = 1929.90
t2 KE2 =(93.24/2)* 25.736^2 = 30878.37
t3 KE3 =(93.24/2)* 54.689^2 =139435.14
etc.
The instantaneous power (P) equals the change in work
(Delta W) done, divided by the change Delta t the time between
samples.
However, the change in work is (Delta W) =(Delta KE)
Therefore the Power (ft lb/sec)= the change in kinetic
energy over the sample time (Delta t).
P = (Delta KE)/(Delta t)
There are 550 ft lb/sec per horsepower therefore
HP = P/550
P1 = (1929.9-0)/1 sec =1929.9 ft lb/sec
or 1929.9/550 = 3.51 HP
P2 =(30878.37-1929.9)/1 sec = 28948.47ftlb/sec 52.63HP
etc.
Time_____KE_____Power_______HP
(sec) (ftlb/sec)
t0_________0.0_______0.0______0.0
t1______1929.9____1929.9______3.51
t2____30878.37___28948.47____52.63
t3___139435.14__108556.77___197.38
t4___351723.93__212288.79___385.98
t5___608038.53__256314.59___466.03
t6___851085.07__243046.54___441.90
t7__1065786.23__214701.17___390.37
t8__1206186.32__140400.09___255.27
t9__1279643.07___73456.75___133.56
t10_1304611.13___24968.06____45.40
t11_1304611.13_______0.00_____0.00
Now here is the problem that I mentioned up front.
After 10 seconds the velocity does not change. Thus the
kinetic energy does not change and stays constant. Thus
the work is zero (W = 0) and the power is zero (P =0).
For example the velocity (v1) at one second (t1) is;
v1=((a0+a1)/2 *Delta t)+a0=((0.00+12.868)/2) *1)+0 = 6.434
v2=((a1+a2)/2 *Delta t)+a1=((12.868+25.736)/2)*1)+6.434=
25.736
etc.
If we plot the velocity as a function of time we obtain
the distance (d)traveled.
For example the distance d1 at t1 is
d1=((v0+v1)/2 *Delta t)+d0=((0.00+6.434)/2)*1)+0 = 3.22
d2=((v1+v2)/2 *Delta t)+d1=((6.434+25.736)/2)*1)+3.22=
19.30
etc.
Time__acceleration____velocity-MPH_distance
(sec__Gs_(ft/sec^2)___(ft/sec)_MPH_____(ft)
t0___0.0____0.00____0.000___0.00____0.00
t1___0.4___12.87____6.434___4.39____3.22
t2___0.8___25.74___25.736__17.55___19.30
t3___1.0___32.17___54.689__37.29___59.51
t4___1.0___32.17___86.859__59.22__130.29
t5___0.7___22.52__114.204__77.87__230.82
t6___0.6___19.30__135.114__92.12__355.48
t7___0.4___12.87__151.199_103.09__498.64
t8___0.2____6.43__160.850_109.67__654.66
t9___0.1____3.22__165.676_112.96__817.92
t10__0.0____0.00__167.284_114.06__984.40
t11__0.0____0.00__167.284_114.06_1151.69
Step 2. Using the velocity and fixed mass (m) of the
car, you then calculate the kinetic energy of the car
as a function of time.
Kinetic energy (KE)equals the car’s mass (m)divided by
2, times the velocity (v)squared (^2))
KE = (m/2)*v^2
t0 KE0 = 0
t1 KE1 =(93.24/2)* 6.434^2 = 1929.90
t2 KE2 =(93.24/2)* 25.736^2 = 30878.37
t3 KE3 =(93.24/2)* 54.689^2 =139435.14
etc.
The instantaneous power (P) equals the change in work
(Delta W) done, divided by the change Delta t the time between
samples.
However, the change in work is (Delta W) =(Delta KE)
Therefore the Power (ft lb/sec)= the change in kinetic
energy over the sample time (Delta t).
P = (Delta KE)/(Delta t)
There are 550 ft lb/sec per horsepower therefore
HP = P/550
P1 = (1929.9-0)/1 sec =1929.9 ft lb/sec
or 1929.9/550 = 3.51 HP
P2 =(30878.37-1929.9)/1 sec = 28948.47ftlb/sec 52.63HP
etc.
Time_____KE_____Power_______HP
(sec) (ftlb/sec)
t0_________0.0_______0.0______0.0
t1______1929.9____1929.9______3.51
t2____30878.37___28948.47____52.63
t3___139435.14__108556.77___197.38
t4___351723.93__212288.79___385.98
t5___608038.53__256314.59___466.03
t6___851085.07__243046.54___441.90
t7__1065786.23__214701.17___390.37
t8__1206186.32__140400.09___255.27
t9__1279643.07___73456.75___133.56
t10_1304611.13___24968.06____45.40
t11_1304611.13_______0.00_____0.00
Now here is the problem that I mentioned up front.
After 10 seconds the velocity does not change. Thus the
kinetic energy does not change and stays constant. Thus
the work is zero (W = 0) and the power is zero (P =0).
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23 June 2006, 07:28 PM
#21
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In answer to your original question - the answer is yes. Problem is that manufacturers' engine-dyno-tested motors will always show less variability in results than rolling-road-tested complete cars once they are out of the showroom and subject to wear, age, modifications from standard etc.
But, all other things being equal, a front wheel drive car will accelerate quicker than a four wheel drive, once you are already rolling and have full traction e.g. between 60-100mph. Of course, all other things are usually NOT equal, but then we get into the realms of Turbo 2000s getting caned around town by hot hatches, or supercars getting caned on long open roads by modified Scoobs etc. etc. All of which only matters if you really care about that kind of thing, and in general the higher the genuine performance of your car, the less you need to flaunt it, hence why I've only ever seen a Ferrari or Lamborghini doing 70-80mph on a motorway and no more. IMHO
But, all other things being equal, a front wheel drive car will accelerate quicker than a four wheel drive, once you are already rolling and have full traction e.g. between 60-100mph. Of course, all other things are usually NOT equal, but then we get into the realms of Turbo 2000s getting caned around town by hot hatches, or supercars getting caned on long open roads by modified Scoobs etc. etc. All of which only matters if you really care about that kind of thing, and in general the higher the genuine performance of your car, the less you need to flaunt it, hence why I've only ever seen a Ferrari or Lamborghini doing 70-80mph on a motorway and no more. IMHO
23 June 2006, 08:16 PM
#22
BHP per tonne is fine, but then you're missing out the torque thump in the midrange from a turbo, and ignoring wind resistane which has a similar effect to adding weight as you add speed.
The faster you go the less difference power/weight makes compared with drag/power.
So you'll find on paper though the lightweight n/a car is faster in real life this is not the case.
The faster you go the less difference power/weight makes compared with drag/power.
So you'll find on paper though the lightweight n/a car is faster in real life this is not the case.
23 June 2006, 09:04 PM
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Try this http://www.letstorquebhp.com/
Click on the link LetsTorqueBHP.com's Performance Calculator On that page. Looks like they have stopped a direct link to it....
Enter the BHP/Kg/Drive and have fun.
Click on the link LetsTorqueBHP.com's Performance Calculator On that page. Looks like they have stopped a direct link to it....
Enter the BHP/Kg/Drive and have fun.
23 June 2006, 10:15 PM
#26
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Originally Posted by Zen Performance
Why not compare actual acceleration, the black dyno never lies 
And don't I know it
Mark.
23 June 2006, 10:46 PM
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Originally Posted by Zen Performance
Why not compare actual acceleration, the black dyno never lies
Check the 60ft times and see how the obviously wrong ones biring the overall time down inline with the 60ft.
Zen, I'd be impressed if even you could do a 1.0 60ft if you don't think it's wrong.
24 June 2006, 10:37 AM
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Originally Posted by Carl2
Yes it does. Proof HERE
Check the 60ft times and see how the obviously wrong ones biring the overall time down inline with the 60ft.
Zen, I'd be impressed if even you could do a 1.0 60ft if you don't think it's wrong.
Check the 60ft times and see how the obviously wrong ones biring the overall time down inline with the 60ft.
Zen, I'd be impressed if even you could do a 1.0 60ft if you don't think it's wrong.
24 June 2006, 01:51 PM
#30
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Originally Posted by Zen Performance
It's all in the terminal...
Terminal certainly shows who has more power, or a higher power to weight ratio, bit it doesn't guarantee who's going to be first across the line !
Mark.