[PRCo] Drum___Brakes

[Jim Holland]

----- Original Message -----
From: "Boris Cefer" <boris6 at volny.cz>
To: <pittsburgh-railways at dementia.org>
Sent: Friday, 12 March, 2004 3:16 AM
Subject: [PRCo] Re: Fineview___PCCs


 > Wasn't the problem of all-electrics in lack of drum brake adjustment? 
Maybe
 > no. The soft adjustment of standard all-electric drum brake would provide
 > very long stopping distance on steeper downgrade and the shoes would 
suffer
 > heavy wear.

 > Maintenance specification we use here for drum brake says that each 
car must
 > be tested by removing of drum brake fuse and pressing power pedal to 
reach
 > starting current of 290 Amps - drum brakes must hold the car (on dead 
level
 > track). 290 Amps cause approximately the same traction effort as 9.5 %
 > downgrade. Sometimes we experience drums which are sufficient to hold the
 > car at 350 Amps, which would be about 11.5 % downgrade, but this is 
with new
 > brake shoes and freshly adjusted brake. I think there is no considerable
 > difference between WAB brake and our CKD product, as for the braking
 > capability.

 > But if you find any drawing and cpecification for WAB drum brake and
 > actuator at Arden (that would need some work - consign it to limbo), 
I can
 > easily made a calculation and answer the question of maximum grade which
 > allows drum brake operation.

Found a *Clark__Equipment__Company* drawing (#703619,) Full Scale, dated 
1949.09.19 and titled:
"""Motor Mounted - Electric Actuated Drive Shaft Brake For All-Electric 
Car"""
Some Very Interesting information is printed on the face of the drawing 
concerning braking pressures with GE Actuator L-6735653 and I quote this 
exactly, line for line:::::::
+++++++++++++++++++++++++++++++++++++
103% Braking with 160# Pull By Solenoid Actuator
Car Weight = 36,000#
36,000 ÷ 4 = 9000# Load Per Axle
Wheel To Rail ƒ = .2
9000# X .2 = 1800# Tan. Force At Rail
1800# X 12.5 = 22,500# Torque At Wheel
22,500 X 6/43 = 3140# Torque At Drum (TRC Specifications)
Drum Radius = 5.5"
3140 ÷ 5.5" = 570# Tan. Force On Drum
ƒ For Lining To Drum = .316
[Therefore] 570 ÷ .316 = 1800# Press On 2 Shoes = 100% Braking
1800 ÷ 2 = 900# Press Per Shoe
Lever Ratio On Shoe 8.5 To 18.125
900 X 8.5 ÷ 18.125 = 422# Pull On Rod
422# + 30# Spring = 452# Req'd. Total Pull "A"
Bell Crank Ratio = 5.625 To 1.9375
452 X 1.9375 ÷ 5.625 = 156# Pull Req'd By
Solenoid Actuator For 100% Braking
160# Pull Available With Actuator Gives 103% Braking
+++++++++++++++++++++++++++++++++++++
It actually seems that the above *Full__Scale* Drawing is reduced in 
size because I also have Clark Drawing #702455 titled:::::::
"""Motor Mounted - Air Applied Spring Released Propeller Shaft Brake 80% 
Braking with 60 Lbs Air""" ---- and this drawing is also Full__Scale but 
considerably larger than the other Full__Scale Drawing!

 >>> From: "Fred Schneider" fschnei at supernet.com 
<mailto:fschnei at supernet.com>

 >>> Regarding variable rate drum brakes: drums were
 >>> generally used only in conjunction with extended
 >>> dynamic brakes in North America,

 > I don't know much about GE, but Westinghouse equipped
 > PRCo 1700s had their drum brake control circuits
 > designed so as to obtain 3 different braking rates.
 > Wiring diagram shows it clearly. And we had the same
 > arrangement on earlier equipment.
 >> Jim Holland wrote:

 >> Are you able to calculate the braking effect and
 >> grades on which the drums would hold the car at these
 >> 3-different rates?

 > I forgot to mention. Each rate (on all-electric) is for one 
particular brake
 > pedal position; in parked position (pedal half way) the drum 
solenoids have
 > no power which means the maximum braking effort.
 > I think if you find drawings with some important parameters, I can 
calculate
 > it. I hope.

 >> Do you think this would be common with all
 >> WABCo actuators or does it also depend upon the drum
 >> brake itself?

 > I don't know details of other actuators. But if the principle of it 
is that
 > braking effect is reached simply mechanical way by means of a spring and
 > releasing is done by a magnetic coil which acts directly against the 
spring,
 > then the actuator can provide particular range of braking effect. But
 > remember that NOT all coils, magnetical cores, main springs and pasive
 > friction between moving parts in mechanism are identical on each 
actuator an
 > d if you give the same lower voltage to all four actuators on a car, the
 > braking effort of each particular brake will slightly differ. And the
 > difference between particular brakes increases with drum shoe wear.




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Jim

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