Mustang Vs. Airboy

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Old Mar 29, 2010 | 11:35 AM
  #16  
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Originally Posted by RussB
wouldn't it be that a dynojet without the RPM pickup only displays HP (i'm assuming that is HP vs. roller rpm?) because it lacks the missing piece of information to calculate the gear reduction, thus not able to calculate "engine" torque? the dyno measures the torque applied to the rollers and the rpm of the rollers and that is enough to calculate HP.

that calculated HP number would be the same regardless of being calculated as:
((roller torque * roller RPM) / 5252)
or as
(((roller torque / gear reduction) * (roller RPM * gear reduction)) / 5252)
The software could calculate torque with respect to roller speed, but that would be pretty useless data. When it lacks an RPM pickup, it just displays HP over time.

The dynojet doesn't actually measure the torque at the rollers. The only data it acquires is the roller speed and time. Then, using the mass of the rollers as calibration data, it can derive the work done over time (engine power) and then it can derive engine torque if engine RPM data is available.

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-- Ed
Old Mar 29, 2010 | 01:36 PM
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Incidentally, Airboy contacted me and is grabbing data from my pulls to investigate his interpolation. His sheet uses linear regressions (as you can see by the last data points)..
Old Mar 29, 2010 | 04:07 PM
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Originally Posted by EQ Tuning
The dynojet doesn't actually measure the torque at the rollers. The only data it acquires is the roller speed and time. Then, using the mass of the rollers as calibration data, it can derive the work done over time (engine power) and then it can derive engine torque if engine RPM data is available.

Thanks
-- Ed
this doesn't seem completely correct either. I could just be confusing myself with all this talk in customary units (all my teachers/profs used traditional (SI) units) and the fact that i have no idea how a dynojet works mechanically...

Why can't the dynojet measure torque directly? knowing the roller RPM, roller mass, and roller radius... don't you have everything you need to solve for t in t=r*F ?
Old Mar 29, 2010 | 04:08 PM
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Originally Posted by Bouncer
Incidentally, Airboy contacted me and is grabbing data from my pulls to investigate his interpolation. His sheet uses linear regressions (as you can see by the last data points)..
well thats cool at least, good to see them trying to improve the product
Old Mar 29, 2010 | 04:56 PM
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Originally Posted by irrational x
well thats cool at least, good to see them trying to improve the product
Agreed. I'm just concerned about stale plots scattered about. It would be nice if an indicator was included in the airboy plots so you know if it's the old calc or updated (assuming an update does happen).
Old Mar 29, 2010 | 05:15 PM
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Originally Posted by irrational x
this doesn't seem completely correct either. I could just be confusing myself with all this talk in customary units (all my teachers/profs used traditional (SI) units) and the fact that i have no idea how a dynojet works mechanically...

Why can't the dynojet measure torque directly? knowing the roller RPM, roller mass, and roller radius... don't you have everything you need to solve for t in t=r*F ?
you'd think so, but you're not a tuner so you don't really know anything...
Old Mar 29, 2010 | 05:38 PM
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Originally Posted by irrational x
this doesn't seem completely correct either. I could just be confusing myself with all this talk in customary units (all my teachers/profs used traditional (SI) units) and the fact that i have no idea how a dynojet works mechanically...

Why can't the dynojet measure torque directly? knowing the roller RPM, roller mass, and roller radius... don't you have everything you need to solve for t in t=r*F ?
You're completely right that given the right variables, the software can calculate and solve for torque. But that doesn't change the fact that the machine is not directly measuring torque to begin with. All its measuring is roller speed and calculating everything else from that data.

Thanks
-- Ed
Old Mar 29, 2010 | 09:40 PM
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Originally Posted by EQ Tuning
You're completely right that given the right variables, the software can calculate and solve for torque. But that doesn't change the fact that the machine is not directly measuring torque to begin with. All its measuring is roller speed and calculating everything else from that data.

Thanks
-- Ed
i guess i'm just not seeing the difference between that and any other physical dyno. does a mustang dyno just measure more inputs like Kw?

paging paul.. i need this translated to retard so i can figure it out. thanks to rick, i know you're fluent.
Old Mar 30, 2010 | 09:31 AM
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Simplified -

Dynojet = inertia dyno. Spins large fixed weight rollers and interpolates power based on acceleration.

Mustang = Brake dyno. Dyno actually applies known load to small rollers via eddy current generator. Using known load applied measures torque.

Dynojet = $$
Mustang = $$$$$$$$$$$$$$$$$$$$$

Totally different category and complexity of the machine itself.

Last edited by Bouncer; Mar 30, 2010 at 09:33 AM.
Old Mar 30, 2010 | 09:35 AM
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For tuning puproses: Mustang > Dynojet
For bragging rights: Dynojet > Mustang

For those who care about bragging rights: Trap Speed >>> Dyno


Edit - Also, for reference.. Engine dynos are almost ALL Brake type dyno's with eddy current generators.
Old Mar 30, 2010 | 09:42 AM
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Originally Posted by Bouncer
Simplified -

Dynojet = inertia dyno. Spins large fixed weight rollers and interpolates power based on acceleration.

Mustang = Brake dyno. Dyno actually applies known load to small rollers via eddy current generator. Using known load applied measures torque.

Dynojet = $$
Mustang = $$$$$$$$$$$$$$$$$$$$$

Totally different category and complexity of the machine itself.
aaah that makes more sense. dynojet measures Δlinear momentum to calc torque while a mustang also measures ΔKw so it can can do a direct conversion to HP.
Old Mar 30, 2010 | 09:49 AM
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Originally Posted by irrational x
aaah that makes more sense. dynojet measures Δlinear momentum to calc torque while a mustang also measures ΔKw so it can can do a direct conversion to HP.
Dynojet = Work/time = Derived unit of HP

Mustang = direct measurement of force
Old Mar 30, 2010 | 09:53 AM
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Now that you've dumbed it down, I understand
Old Mar 30, 2010 | 09:57 AM
  #29  
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Not dumbed it down.. Just explained it more directly.

Mustang uses force and then calculates the derived unit of HP

All in all, the final numbers are calculations and depend heavily on calibration.

You can set the mustang to apply load that is roughly equiv. to the vehicle weight.

In the end they are all simulations of actual road conditions. Your results may vary.
Old Mar 30, 2010 | 10:07 AM
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Originally Posted by Bouncer
Not dumbed it down.. Just explained it more directly.

Mustang uses force and then calculates the derived unit of HP

All in all, the final numbers are calculations and depend heavily on calibration.

You can set the mustang to apply load that is roughly equiv. to the vehicle weight.

In the end they are all simulations of actual road conditions. Your results may vary.

Since F=d/dt(mv) is the mustang using roller mass and rotation to calculate torque and current in Kw to calc HP? or does the eddy current generator create resistance and simulate the roller mass?



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