**Description of events**

**E1:** **S starts the initial acceleration to relative velocity v**

**E1a: S ends the initial acceleration to relative velocity v**

**E2: S starts his turnaround acceleration**

**E2a:** **S is momentarily at rest in R**.

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**E3:** **S ends his turnaround acceleration**

**E4a: S starts the final acceleration from relative velocity v to being at rest with R**

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**E4:** **S ends the final acceleration from relative velocity v to being at rest with R**

**R’s clock records 200 billion ticks between E1 and E4**. **We set v at 0.866c so that the time dilation factor is ½ . Hence, S’s clock records 100 billion ticks between E1 and E4 and the net difference in proper time is 100 billion ticks.** **Since the constant velocity segments can be made arbitrarily long versus the period of acceleration, we choose, for this scenario, that S's clock accumulates 1,000 ticks of proper time between E2 and E3. The proponent of a reconciliation argument is asked to give the number of ticks (i.e., the accumulated proper time) for the traveling twin, S, for the segments defined by the pairs of events E1 - E1a, E1a - E2, E2 - E2a, E2a - E3, E3 - E4a and E4a - E4. The sum should add up to 100 billion**. **For the acceleration segments, the responder may give approximations.** .