## What are the powertrain calculations for the E-Baja competition?

It is similar to m-BAJA powertrain calculation.

First of all, you should find drag forces on the vehicle then find out the torque on wheels, after that decide your reduction on the basis of your motor output torque. Its depend on you that which type of transmission system you choose (auto or manual). Then find your acceleration, top speed and gradeability of vehicle.

For the basics take help from fundamental of vehicle dynamics by Thomas d. Gillespie and for coefficients and constants take help from google search engine.

## Does eBaja ATV transmission require CVT?

E BAJA ATV does not require CVT.

Only Motor and a gearbox do it all of the power transmission to the wheels.

As the voltage varies through the throttle, the motor’s torque and RPM are varied…so no CVT required for variable transmission.

## Manual Transmission calculations – 2 | Tractive forces, accelaration and gradability

Some basic formula for the calucation of various dynamics of vehicle…..for gradability the force component will be resolve in two component one is along the road and one is perpendicular to the road. only along the road component will help in climbing and perpendicular to the road will cause the vehicle to topple. there are various losses in engine , tranmission due to various friction so the result which is gonna come is gonna be in ideal case(i.e no friction etc) so what ever you get after calculation the actual result will be only 60-70 % of that. so calculate accordingly.

here is sample calculation(  ideal case)

At hill climbing, tractive force available at an inclination of 45°,
F=Wsinθ- μWcosθ ≈2081N
Power =10hp=7460 Watt =F*V
V=3.58m/sec=12.9km/hr
So, our vehicle can climb the hill of inclination 45° with this velocity.
Air resistance (Ra)= Ka*A*(V)²
 Ka= 0.004725 – 0.005 ,depending upon type of vehicle we have considered it to be 0.005
 Area (A) = Projected front area= 1.025 m²
 Velocity = 59.06km/hr
Hence Ra=17.88N
Rolling Resistance (Rr) = K*W
 W=total weight of vehicle= 300kg
 K is a constant and depends upon the nature of surface
 We have considered the road surface to be loose mud whose rolling resistance is 0.0435
Hence Rolling resistance (Rr) = 128.02N
Total resistance= Ra + Rr =145.9N
Traction –
F= (Te* Gear ratio* Transmission efficiency)/wheel radius
 Te=engine torque= 19Nm
Hence F=756.21N (at low speed)
Acceleration:
a=(F-R)/m=2.03m/sec²
So, 0-60 km/hr in 8.21 sec.

## Manual Transmission Gearbox – How it Works!

“Automatic vs manual” is a debate that has gone on for years and isn’t coming to a stop any time soon. Even though we don’t see that debate ending, we think that it might be beneficial to drop a little knowledge that could aid in the situation.

Today, we show you what exactly makes a manual transmission tick. What exactly happens when you bang those gears to get your car or truck rolling?

From why the transmission is even necessary to how it takes the power created from the engine and transfers it to the wheels, this demonstration breaks it all down.

Check out the video below to learn a few things about how all of this works. Better break out that notepad and write a few things down because we’re going back to school.

## CALCULATIONS FOR FNR GEAR BOX AND CVT COMBINATION  Gear Reduction Selection

Power Requirements against:

Fr: Wheel Resistance

FL : Air resistance

Fa: Acceleration resistance

Wheel Resistance= Rolling resistance + Slip Resistance + Road Resistance

Tr: Brake Torque

Assumption taken

mass of vehicle being=Mf

CALCULATIONS

Fr=fr*Mf*g*Cosαst     where αst=30˚

Fr=0.16*200*9.81*Cos 30

Fr=280N

R=313*9.81=3070.50N

Static Resistance=µR

=0.65*3070.5=1995.84N

Vmax=60kmph

= .221m

Air Resistance

FL:0.5ρCwAv2

Ρ=Air Density

Cw=Drag Coefficient

A=Area

V=velocity

FL =0.5*1.99*0.0315*(16.66)2

=0.52N

Fst=Mf*g *Sinαst

Putting the corresponding value

Fst=131.725N

Acceleration resistance:

Fa=m*a

=956N

Ftotal=2662.247

Power Required: Ftotal*v/(Nrpm*η)