Quit showing dyno reports as WTQ, because it's wrong and annoying.
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Honda-Tech Member
Joined: Sep 2007
Posts: 55
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From: Tacoma, WA
Quit showing dyno reports as WTQ, because it's wrong and annoying.
It gets on my nerves when I see "175whp/170wtq" in people's sigs because it's just plain wrong. 
It's wrong, because the torque at the wheels depends on your drive train and even in top gear, it's almost ALWAYS greater than the engine and unless you're talking about bicycles.
It is correctly reported as 175fwhp, ~170 lbs-ft. "~" because it's an estimation.
Technical blahs:
Output RPM = input RPM/total gearing
Output torque = input RPM * total gearing
rotational speed * torque * k = power. k=constant, depending on the units.
k is 1.904x10^-4 if the torque is in lbs-ft, rotational speed in RPM and power in hp.
Dyno knows exactly how much rotational inertia its rollers have, so it records the rotational speed of the rollers many times a second, the computer does the integration, then reports the result on the screen.
Since you're not interested in hp with respect to time, the ignition pickup is used to monitor the RPM, so it can draw the hp with respect to engine RPM instead.
If you know the power and RPM, you can calculate torque, but dyno knows nothing about your wheel's RPM since it doesn't know your wheel diameter.
The dyno knows your wheel hp and engine RPM with respect to time, so it takes the (wheel hp - assumed loss) and engine RPM to estimate crank torque. It's only an estimate, because the dyno doesn't know the EXACT drive-train loss, so it uses a generic derating constant.
If your drive-train is more efficient than the assumed constant, it over-reports the torque and vice versa.
/rant
It's wrong, because the torque at the wheels depends on your drive train and even in top gear, it's almost ALWAYS greater than the engine and unless you're talking about bicycles.
It is correctly reported as 175fwhp, ~170 lbs-ft. "~" because it's an estimation.
Technical blahs:
Output RPM = input RPM/total gearing
Output torque = input RPM * total gearing
rotational speed * torque * k = power. k=constant, depending on the units.
k is 1.904x10^-4 if the torque is in lbs-ft, rotational speed in RPM and power in hp.
Dyno knows exactly how much rotational inertia its rollers have, so it records the rotational speed of the rollers many times a second, the computer does the integration, then reports the result on the screen.
Since you're not interested in hp with respect to time, the ignition pickup is used to monitor the RPM, so it can draw the hp with respect to engine RPM instead.
If you know the power and RPM, you can calculate torque, but dyno knows nothing about your wheel's RPM since it doesn't know your wheel diameter.
The dyno knows your wheel hp and engine RPM with respect to time, so it takes the (wheel hp - assumed loss) and engine RPM to estimate crank torque. It's only an estimate, because the dyno doesn't know the EXACT drive-train loss, so it uses a generic derating constant.
If your drive-train is more efficient than the assumed constant, it over-reports the torque and vice versa.
/rant
Thread Starter
Honda-Tech Member
Joined: Sep 2007
Posts: 55
Likes: 0
From: Tacoma, WA
Re: Quit showing dyno reports as WTQ, because it's wrong and annoying. (rico91stang)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rico91stang »</TD></TR><TR><TD CLASS="quote">wow. Why does it matter that much. Thanx for the education...I guess.
</TD></TR></TABLE>
Because it's wrong
</TD></TR></TABLE>
Because it's wrong
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