Sbp Gear Speed Calculations
Let's put this in terms of usable math.
Let's say that we have two gears in mesh.
Gear 1 (we'll call it the driver) is turning at speed S1 rpm and has T1 teeth.
Gear 2 (the driven gear) is turning at speed S2 and has T2 teeth.
Then our relationship above says that: ( S1 * T1 = S2 * T2 )
MOTOR REDUCTION BUILT IN
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20 tooth gear 8000 rpm (S1)
82 tooth gear 1951.22 rpm (S2)
ratio 4.10:1
S1= 8000 RPM
T1= 20
S2= 1951.22 RPM
T2= 82
(S2= ?????)
S2 = (T1/T2) * S1 =
((20/82) * 8000) = 1951.22 rpm
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MOTOR OUTPUT 10T SPROCKET
82 tooth gear 1951.22 rpm (S2)
10 tooth gear 1951.22 rpm. (S3)
ratio 0.12:1
S1= 8000 RPM
T1= 20
S2= 1951.22 RPM
T2= 82
S3= 1951.22
T3= 10
SG= 1951.22
S1 * T1 = Sg * T2
Sg = (T1/T2) * S1
8000 x 20 = 1951.22 * 82
SG= ((20/82) x 8000)
*** MAJOR CAVEAT ***
Note that everything said to this point
assumes that each of the gears in the gear train is on its own, separate shaft.
Sometimes gears are 'ganged' by keying or otherwise
and both gears turn as a unit on the same shaft.
This complicates the computation of the gear ratio, but not horribly.
Suppose gears 2 and 3 are keyed together into a single compound gear
we'll designate g (g for ganged).
Assuming S1 and S2 are in mesh, it's still true that:
S1 * T1 = Sg * T2
Sg = (T1/T2) * S1
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LEFT JACKSHAFT SPROCKET 17T
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10 TOOTH SPROCKET 1951.22 RPM (S3)
17 TOOTH SPROCKET 1147.78 RPM (S4)
S1= 8000 RPM
T1= 20
S2= 1951.22 RPM
T2= 82
S3= 1951.22
T3= 10
SG= 1951.22
S4= 1147.78 RPM
T4= 17
S4= ((10/17) x 1951.22) = 1147.78 RPM
S4 = (T3/T4) * Sg = (T3/T4)*(T1/T2) * S1
If gears 3 and 4 are in mesh,
Sg * T3 = S4 * T4
(Remember, S3 turns at the same speed as S2 because they're physically
joined and we're calling their shared speed Sg.)
Therefore,
S4 = (T3/T4) * Sg = (T3/T4)*(T1/T2) * S1
So the end-to-end gear ratio is (T1*T3)/(T2*T4)
it *does* depend on the intermediate gears,
unlike the previous case when each gear could turn on its own separate axis.
Note that the resultant gear ratio
is just the product of the two separate gear ratios - (T1/T2)*(T3/T4).
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RIGHT JACKSHAFT SPROCKET 17T
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17 TOOTH SPROCKET 1147.78 RPM (S4)
11 TOOTH SPROCKET 1147.78 RPM (S5)
S1= 8000 RPM
T1= 20
S2= 1951.22 RPM
T2= 82
S3= 1951.22
T3= 10
SG= 1951.22
S4= 1147.78 RPM
T4= 17
S5= 1147.78 RPM
T5= 11
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FREEWHEEL CHAINRING 48T
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11 TOOTH SPROCKET 1147.78 RPM (S5)
48 TOOTH CHAINRING 263.03 RPM (S6)
S5= 1147.78 RPM
T5= 11
S6= 263.03 RPM
T6= 48
(S6= 263.03)
S6 = (T5/T6) * S5 =
((11/48) * 1147.78) = 263.03 rpm
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FREEWHEEL CHAINRING 36T
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48 TOOTH CHAINRING 263.03 RPM (S6)
36 TOOTH CHAINRING 263.03 RPM (S7)
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REAR SPROCKET 32T
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36 TOOTH CHAINRING 263.03 RPM (S7)
32 TOOTH CHAINRING 213.71 RPM (S8)
S7= 263.03 RPM
T7= 36
S8= 213.71 RPM
T8= 32
(S8= 263.03)
S8 = (T7/T8) * S7 = 213.71 RPM
((36/32) * 263.03) = 213.71 rpm
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REAR SPROCKET 11T
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36 TOOTH CHAINRING 263.03 RPM (S7)
11 TOOTH SPROCKET 860.825 RPM (S9)
S7= 213.71 RPM
T7= 36
S9= 860.825
T9= 11
(S9= 860.825)
S9 = (T7/T9) * S7 = 860.825 RPM
((36/11) * 263.03) = 860.825 rpm
NOW WE KNOW THE SPEED OF EACH GEAR.
MUCH NEEDED INFO TO SELECT PROPER BEARINGS.
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EST MPH @ EXACT 26" DIA WHEEL.
Calculate the circumference of a circle:
C = 2 * Π * r
C = 81.68141 inches
FEET IN A MILE = 5280
5280 x 12 = 63360
63360 / 81.68141 = 775.69 ROTATIONS PER MILE
213.71 X 60 = 12822.6 ROTATIONS PER HR
12822.6 / 775.69 = 16.53 MPH 1ST GEAR @ 8000 RPM MOTOR
860.825 X 60 = 51649.5 ROTATIONS PER HR
51649.5 / 775.69 = 66.58 MPH HIGH GEAR @ 8000 RPM MOTOR
WOW THATS FAST.
ID HATE TO HAVE MY FREEWHEEL FAIL AT THAT SPEED.
YOU PEDALS WOULD BE MOVIN AT 263.03 RPM
I'VE HEARD THE STANDARD IS 100 RPM FOR A HUMAN