honda cbr torque values

[aious]

wat is the approximate measured torque and power value for honda cbr 600 cc f4i

[aious]

wat is the approximate measured torque and power value for honda cbr 600 cc f4i

[benjo]

Google is your friend...

First two results gave me torque figures of about 65 Nm stock.

[exFSAE]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by benjo:
Google is your friend...

First two results gave me torque figures of about 65 Nm stock. </div></BLOCKQUOTE>

Not sure what this "Newton meters" sillyness is about.

IIRC, should be about 0.25 slug-force fathoms peak output torque.

[The AFX Master]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by exFSAE:
Not sure what this "Newton meters" sillyness is about.
IIRC, should be about 0.25 slug-force fathoms peak output torque. </div></BLOCKQUOTE>

Dude.. not to sound trollish, but... why always your comments are useful and 75% BS??.
Who cares about what units??, do useless comments and don´t be a forum troll.

Back in topic, on the web you´ll find torque curves from the stock bike, unrestricted and with another ECU mapping and exhaust. that are far from our operating conditions
Do a search here on the forum for FSAE like curves.

[exFSAE]

Nah mate my comments are only 72% BS. Can't believe you'd make such an outlandish statement that I am all the way up to 75.

For reals. No BS. 1/4 slug-fathom is right in ballpark for output torque. Sarcasm? Me? Absolutely. More silly the thread, more sarcastic I am.

I, like others, really dislike questions here that are outrageously simple. Eg stuff that should (a) be already covered in frosh or soph level engineering classes or (b) be covered in a 30 second google search.

When people pose legitimate questions or ideas, I respond accordingly.

[PatClarke]

Aious,
After the responses you got on your first post, I would have thought you understood that the population here require posters to do some work themselves.
I advised you to use Google. My Google research answered both your questions in the first couple of lines. That shows me you are lazy!

If it is your intent to enter FSAE as part of building a career for yourself after University, then I suggest you need to become far more self sufficient. Noone is going to give you anything in the hard world of business.

Now, if you intend to enter FSAE and you intend to use a CBR600, then a little research would have shown you that many winning teams use this engine. Your project will not depend on the amount of power or torque you imagine you can get, but rather that you get to the competition with a complete car that reflects your abilities. Go do it!

Finally, there are 2 'sticky' threads on here you should read and understand.
One is for new posters advising them to use the search function. The other is specifically for Indian teams.
Read them both!

The information held in the archives at this site is the best resource available on the subject. It is available free for anyone prepared to do some work searching, so don't be lazy...Start searching!

Pat

[Nishant Jain]

Hey, seeing as how this thread is on about Honda engine and torque values...

I wanted some help in calculating, or rather, verifying the safety in the shaft diameter we have chosen. Ours is a 20mm dia rear axle shaft. To make fatigue shear stress calculations on it, I have a value of final torque at the rear axle after all the gearing from the crankshaft onwards. Of course, I also have the shaft material properties so I know the shear strength. Since we haven't been running this car for too long now, we dont really know the variations in value of the torque whilst the car runs.

For fatigue analysis, I need the mean torque value while the car runs as well as the amplitude variation in this torque value.

Now, considering a normal track run, is it right if I assume the car to be running on 2nd gear primarily 80% close to the red line rpm? This way I can have a torque at that point. Then, again I'm guessing, a variation of 10% on either side of this torque figure?

Is that okay? It's kind of difficult to reasonably know these figures without having a car to run from before.

[Pete M]

I'd say think about what happens when the driver lifts off the throttle completely and you get engine braking.

[Nishant Jain]

@Pete: Ok. I have a value for the torque applied on the rear axle due to braking. That would be my lower value of torque then?

I suppose then the braking torque is minimum value while the engine torque would be the maximum. Is that correct?

Hmm. The g-g diagram in RCVD (Milliken and Milliken)?

[Superfast Matt McCoy]

This question doesn't really have anything to do with the engine or its torque values. The maximum torque your axles will see is based on the tires. You need to calculate the maximum torque the axles see before the tires break loose based on the friction of the tires, which will need to take into account the normal force on the tires from the vehicle weight including longitudinal load transfer.

factor in an impact factor (a big one) and a safety factor and you have your maximum torque for your fatigue limit.

It always comes back to the tires.

[murpia]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Superfast Matt McCoy:
This question doesn't really have anything to do with the engine or its torque values. The maximum torque your axles will see is based on the tires.
...
It always comes back to the tires. </div></BLOCKQUOTE>
What about rear wheel inertia? Imagine the engine attempting to drive your wheels at it's maximum torque. Regardless of tyre grip the rear wheel inertia will react against the engine torque and your driveshafts will see full engine torque (multiplied by gear ratio of course).

If you prefer, just imagine driving a pair of remote flywheels with the engine. How strong would the driveshafts to the flywheels need to be?

Regards, Ian

[Superfast Matt McCoy]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by murpia:

What about rear wheel inertia? Imagine the engine attempting to drive your wheels at it's maximum torque. Regardless of tyre grip the rear wheel inertia will react against the engine torque and your driveshafts will see full engine torque (multiplied by gear ratio of course).

If you prefer, just imagine driving a pair of remote flywheels with the engine. How strong would the driveshafts to the flywheels need to be?

Regards, Ian </div></BLOCKQUOTE>

I still say it's irrespective of the engine horsepower. If you're talking about the extra torque to overcome the inertia in an accelerating car, I have to think that happens relatively slowly for the inertia of the drivetrain to encompass a significant amount of the power output.

If you are talking about dumping the clutch at redline, revving at 14k only requires the horsepower to overcome friction and pumping losses. If you dump the clutch, the engine doesn't instantly start making peak torque. Although the axles would likely see a huge torque, that torque would have more to do with the rotating inertia of the engine components than the power output. That's where the impact factor comes in.

[Nishant Jain]

@matt: So if I assume the braking torque (from the friction factor on the rear tyre patch) as my maximum effective torque on the drive shaft, what kind of variations about this value do I consider for fatigue analysis? Or can I just take this number and calculate with this?

I was thinking that the mean torque would be between this brake torque and the engine's driving torque through the drive sprocket. Then I'd have amplitude variations and a mean value to get fatigue analysis on. I guess that's completely wrong?

@murpia: I dont think I completely understand, but are you saying what I am?


And what, please, is "dumping" the clutch???

[murpia]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Superfast Matt McCoy:
I still say it's irrespective of the engine horsepower. If you're talking about the extra torque to overcome the inertia in an accelerating car, I have to think that happens relatively slowly for the inertia of the drivetrain to encompass a significant amount of the power output. </div></BLOCKQUOTE>
Imagine driving in 1st gear at an rpm say 500rpm below where peak engine torque is. You might be at about 15% throttle to maintain constant speed. Now go to 100% throttle - manifold filling effects and transient fuelling will mean a delay in achieving max torque, but you'll get it in under a second probably, certainly within two. Lets assume you reach full torque just as you reach the rpm for max torque. A percentage of that max engine torque will be accelerating the car, the rest will be accelerating the drivetrain.

If you have sufficiently accurate inertia numbers for your car, you could calculate the exact percentage of the torque that's accelerating the crank, gearbox and final drive and subtract that from the total. The rest is accelerating the inertia of the rear wheels, through the driveshafts. It's certainly more than the contact patch forces would suggest.

Or are you suggesting the engine 'knows' you are wheelspinning and somehow decides to produce less torque in this case?

Regards, Ian

[Superfast Matt McCoy]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by murpia:

Imagine driving in 1st gear at an rpm say 500rpm below where peak engine torque is. You might be at about 15% throttle to maintain constant speed. Now go to 100% throttle - manifold filling effects and transient fuelling will mean a delay in achieving max torque, but you'll get it in under a second probably, certainly within two. Lets assume you reach full torque just as you reach the rpm for max torque. A percentage of that max engine torque will be accelerating the car, the rest will be accelerating the drivetrain.

If you have sufficiently accurate inertia numbers for your car, you could calculate the exact percentage of the torque that's accelerating the crank, gearbox and final drive and subtract that from the total. The rest is accelerating the inertia of the rear wheels, through the driveshafts. It's certainly more than the contact patch forces would suggest.

Or are you suggesting the engine 'knows' you are wheelspinning and somehow decides to produce less torque in this case?

Regards, Ian </div></BLOCKQUOTE>

Yea, that does make sense.

It seems like the power required to accelerate everything before the shafts (crank, transmission main and countershafts with gears, chain, sprockets, and diff) would take a large part of the power produced by the engine (or at least a large part of the portion of the power that is going to overcome rotational inertia), and so the shafts would never see full engine torque.

You could lift the rear of the car and stab the throttle, then do the same with the shafts off and look at the difference in rpm vs time slope.

[Nishant Jain]

Hmm. I'm still unsure about what most of this means.

Can you elaborate a little? Also, I'm unfamiliar with the exact jargon you're using.