A servo of this size, stall torque and max speed are all variables, but the half-bridge driver Ron and Motor Coil DCRs.
Ultimately determine peak current \$I_{bat}= \dfrac{(V_{bat} - V_{emf})}{(R_{on} + DCR_{coil})}\$ for the duration of time required to reach max speed then reverse current at speed might be greater with added BEMF voltage at speed velocity. This "start current" is the same as "Stall current" except it reduces as motor ramps up in speed generating -Vemf. That is it is not stalled and thus current ramps down with rising speed towards the load current.
Stall/Start current is typically 10x rated current for 100% duty of a motor, so a battery RMS current and energy Ah depends on the work being done by the servo.
When a digital command to change positions, Newton's Law for acceleration mass and slew time are limited by the torque and speed curve. The gear motor with servo feedback responds to angular position error from initial position to new command position.
I expect the or random seek or average seek over the entire range is equal to 1/3 max stroke. This is the same as in disk drive servos, however I the maximum rotational velocity depends on the load and it only reaches it with no load.
The 2nd Servo does have stall current specs ( error in your links)
No. 工作电压 Operating Voltage Range 4.8V 6V
4-1* 静态电流 Idle current(at stopped) 5mA 6mA
4-2* 空载速度 No load speed 0.13sec/60deg 0.11sec/60deg
4-3* 空载电流 Runnig current(at no load) 130mA 150mA
4-4 堵转扭矩 Peak stall torque 3kg.cm 3.5kg.cm
4-5 堵转电流 **Stall current 700 mA 800 mA**
Conclusions
Estimate: DSM44 servo Isc = 1A @ 6V +/-50% ... x6 motors= 6 A
Specs: FT1117M servo Isc= 800 mA @6V ...... x2 motors = 1.6A for a worst case 7.6A
If 6V supply has an ESR of 50 mΩ then 7.6A x 50 mΩ= 0.38V drop from 6V.
If a supercap was added to a DCDC supply, for a 100 ms seek using an average seek.
hypothetical problem....
The equivalent capacitance for a 0.6V drop from 6V with 6A $$_C=\frac{I_c*dt}{dV}=6A/0.6V * 0.1s= 1 Farad$$