Thanks.

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SCOOP
oddly obsessed with big scoops on little Mustangs

"Why do people who know the least know it the
loudest?"

2.26 60'S
14.9 @ 90.86MPH

65 coupe,351w,C4,Big Boss 429 hood scoop,8"3.40 TracLoc,Cragar SS,Black Suede.

http://www.picturelist.com/auctions/showimagez.asp?user=dbhard&pic=KK333.jpg

Erik

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1966 Ford Mustang 2+2
Mine - Restomod in Progress

Uh oh. Dynojets and high stall converters don't necessarily get along
well. Dynojets work by measuring angular acceleration and the computing
torque. From torque and RPM, they can the compute horsepower:

TQ = d/dt (I * V)

Where:

TQ = torque
I = inertia
V = angular velocity
d/dt = time varying derivative

For a constant inertia and a constant angular acceleration this can be
reduced to:

TQ = I * W

Where:

W = angular acceleration

This is just a rotary equivalent of Newton's Second Law of Motion (the sum
of the external forces acting on a body is equal to the rate of change of
momentum of the body) which reduces to the famous F = M*A (for constant
mass and acceleration).

Since, for an engine, horsepower is the rate of producing torque, we can
convert between these two quantities given the engine rate (RPM):

HP = (TQ*2.0*PI*RPM)/33000.0
TQ = (33000.0*HP)/(2.0*PI*RPM)

where:

TQ = torque in ft-lbs
HP = power in horsepower
RPM = engine speed in revolutions per minute
PI = the mathematical constant PI (approximately 3.141592654)
Note: 33000 = conversion factor (550 ft-lbs/sec * 60 sec/min)

Since the Dynojet drum is of a known inertia and you can measure the angular
acceleration (if you know angular velocity versus time), you can compute the
torque. This works well for a stick shift car since the engine RPM is
proportional to tire RPM which turns the drum. However, for an automatic
there is slippage so there isn't a constant relationship between drum RPM
and engine RPM. This only gets worse with a high stall converter. This will
fool the dynojet software and lead to false torque readings.